Understanding Amount of Substance and the Mole: A Comprehensive Guide

Introduction to the Concept of Amount of Substance

In the world of chemistry, the amount of substance is a fundamental concept that allows us to quantify the number of particles (such as atoms, molecules, or ions) in a sample. Whether you're studying chemical reactions, stoichiometry, or concentrations, understanding the amount of substance helps connect the microscopic world to the macroscopic, observable world.

Amount of substance is typically measured in moles, which makes it essential for both academic studies and real-world applications. In this guide, we will explore the significance of moles, their relationship with mass and volume, and how to use the concept in various calculations.

What is the Mole in Chemistry?

The mole (symbol: mol) is the standard SI unit used to measure the amount of substance. A mole of any substance contains Avogadro's number of particles, which is approximately 6.022 × 10²³. This number represents the number of particles in one mole of a substance, whether they are atoms, molecules, or ions.

Why is the Mole Important in Chemistry?

The mole concept allows chemists to translate between the atomic scale and the macroscopic scale. Without the mole, the vast number of atoms or molecules in even a small sample of matter would be difficult to handle. By using the mole, chemists can work with manageable numbers and predict quantities involved in chemical reactions with precision.

Understanding Avogadro's Number: The Foundation of the Mole

Avogadro's number ($6.022 \times 10^{23}$) is a constant that defines the number of atoms, molecules, or ions in one mole of a substance. This constant plays a vital role in chemistry because it provides a bridge between the atomic scale and our everyday scale.

For instance, if you have a mole of water molecules (H₂O), it will contain 6.022 × 10²³ molecules of water, which corresponds to approximately 18 grams of water. This allows chemists to easily calculate the amount of substance in grams, moles, or molecules.

Molar Mass: The Key to Converting Between Moles and Mass

Molar mass is defined as the mass of one mole of a substance, and it is usually expressed in grams per mole (g/mol). The molar mass of an element is numerically equivalent to its atomic mass in atomic mass units (amu), and for compounds, it is the sum of the molar masses of the constituent elements.

How to Calculate Molar Mass:

1. Identify the elements in the compound and their atomic masses (from the periodic table).
2. Multiply the atomic mass of each element by the number of atoms of that element in the compound.
3. Add up the total mass for all elements.

Example: The molar mass of water (H₂O):

• Hydrogen (H): 1.008 g/mol × 2 = 2.016 g/mol
• Oxygen (O): 15.999 g/mol
• Total molar mass of H₂O = 18.015 g/mol

Thus, one mole of water weighs 18.015 grams.

Converting Between Moles, Mass, and Number of Particles

Once you understand molar mass, it becomes easy to convert between mass, moles, and the number of particles in a sample. Here are the key conversion formulas:

• Moles = Mass / Molar Mass
• Number of particles = Moles × Avogadro’s number

1. If you have 36.03 grams of water (H₂O), how many moles does this represent?

1. Molar mass of H₂O = 18.015 g/mol
2. Moles = 36.03 g ÷ 18.015 g/mol = 2 moles of H₂O

2. If you have 2 moles of H₂O, how many molecules do you have?

1. Number of molecules = 2 moles × $6.022 \times 10^{23}$ molecules/mol
2. Number of molecules = $1.204 \times 10^{24}$ molecules  

Sample Questions and Answers on Amount of Substance and the Mole

Question 1:

The atomic mass unit (amu) of carbon-12 is $1.6605 \times 10^{-24} \, \text{g}$. If the mass of an atom is $5.313\times 10^{-23}\text{g, determine the relative mass of the atom.}$

Solution:

The relative mass of an atom is calculated as the ratio of the mass of the atom to the atomic mass unit:

$$\text{Relative Mass}=\frac{\text{Mass of the atom}}{\text{Atomic Mass Unit (amu)}}$$

Substituting the given values:

• Mass of the atom: $5.313 \times 10^{-23} \, \text{g}$
  
• Atomic mass unit: $1.6605 \times 10^{-24} \, \text{g}$
  

$$\text{Relative Mass}=\frac{5.313\times 10^{-23}}{1.6605\times 10^{-\mathrm{24}}}$$

Perform the division:

Divide the coefficients:

$$\frac{5.313}{1.6605}\approx 3.2$$

Subtract the exponents of 10:

$$10^{-23}-10^{-24}=10^{-23+24}=10^1$$

Combine the results:

$$3.2\times 10^1=\mathrm{<b>32</b>}$$

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Question 2:

How many atoms are there in 0.3 mol of sodium? Given $L = 6.0 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's constant).

Solution:

The number of atoms in a substance can be calculated using the formula:

$$\text{Number of atoms}=\text{moles}\times L$$

Where:

• $\text{moles}=0.3\text{mol}$
• $L = 6.0 \times 10^{23} \, \text{mol}^{-1}$
  

Substituting the values:

$$\text{Number of atoms}=0.3\times 6.0\times 10^{23}$$

Perform the calculation:

1. Multiply the coefficients:

   $$0.3\times 6.0=1.8$$

2. Include the power of 10:

   $$1.8\times 10^{23}$$

The number of atoms in $0.3 \, \text{mol}$ of sodium is $1.8\times 10^{23}\text{atoms.}$

---

Question 3:

Calculate the amount of oxygen in moles in $1.505 \times 10^{23}$ molecules of oxygen gas. Given $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's constant).

Solution:

The number of moles can be calculated using the formula:

$$\text{Moles} = \frac{\text{Number of molecules}}{\text{Avogadro's constant (L)}}$$

Where:

• $\text{Number of molecules} = 1.505 \times 10^{23}$
  
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$
  

Substituting the values:

$$\text{Moles} = \frac{1.505 \times 10^{23}}{6.02 \times 10^{23}}$$

Perform the calculation:

1. Divide the coefficients:

   $$\frac{1.505}{6.02} \approx 0.25$$
   

2. Subtract the exponents of $10^{23}$:

   $$10^{23} - 10^{23} = 10^0 = 1$$
   

3. Combine the results:

   $$0.25 \times 1 = 0.25$$The amount of oxygen is 0.25 moles.

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Question 4:

Calculate the number of oxygen atoms in $0.20 \, \text{mol}$ of oxygen gas. Given $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$.

Solution:

Step 1: Relationship between moles of $O_2$ molecules and oxygen atoms

Oxygen gas exists as diatomic molecules ($O_2$), so, each molecule contains 2 oxygen atoms. The total number of oxygen atoms can be found using:

$$\text{Number of atoms}=\text{moles of }{O}_2\times L\times 2$$

Where:

• $\text{moles of } O_2 = 0.20 \, \text{mol}$
  
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$
  
• $2 = \text{atoms per molecule of } O_2$

Step 2: Substitute the values

$$\text{Number of atoms}=0.20\times 6.02\times 10^{23}\times 2$$

Step 3: Perform the calculation

1. Multiply the coefficients:

   $$0.20\times 6.02=1.204$$

2. Multiply by 2:

   $$1.204 \times 2 = 2.408$$
   

3. Include the power of 10:

   $$2.408 \times 10^{23}$$

   The number of oxygen atoms in $0.20\text{mol}$ of oxygen is $\mathrm{<b>2.408</b>}<b>\times </b>\mathrm{<b>1</b>}{\mathrm{<b>0</b>}}^{\mathrm{<b>23</b>}}\text{<b>\hspace{0.17em}</b>}\text{atoms}$.

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Question 5:

Calculate:

(i)  the number of molecules contained in $0.6 \, \text{mol}$ of $\text{H}_2\text{S}$.
(ii) the number of atoms of H and S present in $0.6 \, \text{mol}$ of $\text{H}_2\text{S}$.

Solution:

Part (i): 

The number of molecules in a substance is calculated using the formula:

$$\text{Number of molecules}=\text{moles}\times L$$

Where:

• $\text{moles} = 0.6 \, \text{mol}$
  
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's constant)

Substituting the values:

$$\text{Number of molecules}=0.6\times 6.02\times 10^{23}$$

Perform the calculation:

$$0.6\times 6.02=3.612$$

Include the power of $10^{23}$:

$$\text{Number of molecules}=3.612\times 10^{23}$$

Answer for (i): The number of $\text{H}_2\text{S}$ molecules is $3.612 \times 10^{23}$.

Part (ii): 

1. Atoms per molecule of $\text{H}_2\text{S}$:

   • Each $\text{H}_2\text{S}$ molecule contains:
     • 2 hydrogen ($H$) atoms
     • 1 sulfur ($S$) atom

2. Total number of atoms of each type:

   • Hydrogen atoms:

     $$\text{Number of H atoms}=\text{Number of molecules}\times 2$$

     Substituting:

     $$\text{Number of H atoms}=3.612\times 10^{23}\times 2=7.224\times 10^{23}$$

   • Sulfur atoms:

     $$\text{Number of S atoms} = \text{Number of molecules} = 3.612 \times 10^{23}$$
     

Final Answer:

(i) The number of $\text{H}_2\text{S}$ molecules is $3.612 \times 10^{23}$.
(ii) The number of hydrogen ($H$) atoms is $7.224 \times 10^{23}$, and the number of sulfur ($S$) atoms is $3.612 \times 10^{23}$.

---

Question 6:

Calculate the number of nitrogen ($N$) atoms in $0.25 \, \text{mol}$ of nitrogen (I) oxide ($\text{N}_2\text{O}$). Given $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$.

Solution:

Step 1: Relationship between moles, molecules, and atoms

Nitrogen (I) oxide ($\text{N}_2\text{O}$) contains 2 nitrogen atoms per molecule.

The total number of $N$ atoms can be calculated using:

$$\text{Number of atoms of }N=\text{moles of }{\text{N}}_2\text{O}\times L\times \text{number of }N\text{ atoms per molecule}\mathrm{.}$$

Step 2: Substituting the values

• Moles of ${\text{N}}_2\text{O}=0.25\text{mol}$
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$
  
• Number of $N$ atoms per molecule = 2

$$\text{Number of atoms of } N = 0.25 \times 6.02 \times 10^{23} \times 2$$

Step 3: Perform the calculation

1. Multiply the coefficients:

   $$0.25 \times 6.02 = 1.505$$
   $$1.505 \times 2 = 3.01$$
   

2. Include the power of $10^{23}$:

   $$3.01 \times 10^{23}$$
   

Final Answer:

The number of nitrogen ($N$) atoms in $0.25\text{mol of }$$\text{N}_2\text{O}$ is $3.01 \times 10^{23} \, \text{atoms}$.

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Question 7:

Calculate the mass of $0.2 \, \text{mol}$ of magnesium atoms. Given the molar mass of magnesium ($\text{Mg}$) = $24\text{g/mol.}$

Solution:

The mass of a substance can be calculated using the formula:

$$\text{Mass}=\text{Moles}\times \text{Molar Mass}$$

Where:

• $\text{Moles} = 0.2 \, \text{mol}$
  
• $\text{Molar Mass} = 24 \, \text{g/mol}$
  

Substituting the values:

$$\text{Mass}=0.2\times 24$$

Perform the calculation:

$$\text{Mass} = 4.8 \, \text{g}$$

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Question 8:

Calculate the molar mass of carbon (IV) oxide ($\text{CO}_2$) if $0.25 \, \text{mol}$ has a mass of $11 \, \text{g}$.

Solution:

The molar mass ($M$) of a substance can be calculated using the formula:

$$M=\frac{\text{Mass}}{\text{Moles}}$$

Where:

• $\text{Mass} = 11 \, \text{g}$
  
• $\text{Moles} = 0.25 \, \text{mol}$
  

Substituting the values:

$$M = \frac{11}{0.25}$$

Perform the calculation:

$$M = 44 \, \text{g/mol}$$

Final Answer:

The molar mass of carbon (IV) oxide ($\text{CO}_2$) is 44 g/mol.

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Question 9:

Calculate the amount of oxygen in moles in $8 \, \text{g}$ of oxygen gas ($O_2$).

Solution:

The number of moles is calculated using the formula:

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$

Step 1: Determine the molar mass of oxygen gas ($O_2$) 

The atomic mass of oxygen ($O$) = $16 \, \text{g/mol}$. Since oxygen gas is diatomic ($O_2$):

$$\text{Molar Mass of } O_2 = 16 \times 2 = 32 \, \text{g/mol}.$$

Step 2: Substitute the values into the formula:

• $\text{Mass} = 8 \, \text{g}$
  
• $\text{Molar Mass of } O_2 = 32 \, \text{g/mol}$
  

$$\text{Moles} = \frac{8}{32}$$

Step 3: Perform the calculation:

$$\text{Moles} = 0.25 \, \text{mol}.$$

Final Answer:

The amount of oxygen gas in $8 \, \text{g}$ is 0.25 moles.

---

Question 10:

What mass of sodium is contained in $53\text{g of sodium carbonate (}$$\text{Na}_2\text{CO}_3$)? Given that [$\text{Na}=23,\text{O}=16\text{<br>},\text{C}=12$].

Solution:

Step 1: Calculate the molar mass of sodium carbonate 

The molar mass of $\text{Na}_2\text{CO}_3$ can be found by adding the atomic masses of its components:

$$\text{Molar Mass of } \text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16)$$
$$\text{Molar Mass of } \text{Na}_2\text{CO}_3 = 46 + 12 + 48 = 106 \, \text{g/mol}$$

Step 2: Determine the mass of sodium in

There are 2 sodium atoms in each molecule of sodium carbonate, and the molar mass of sodium is $23 \, \text{g/mol}$. So, the mass of sodium in one mole of $\text{Na}_2\text{CO}_3$ is:

$$\text{Mass of Sodium}=2\times 23=46\text{g}\mathrm{.}$$

Step 3: Use the proportion to find the mass of sodium in $53\text{g}$ of${\mathrm{ <span\; style="font-family:\; "Times\; New\; Roman";\; font-size:\; medium;"> </span>}}_{}$${\text{Na}}_2{\text{CO}}_3$

We know that $106 \, \text{g}$ of $\text{Na}_2\text{CO}_3$ contains $46 \, \text{g}$ of sodium. We can set up the following proportion:

$$\frac{\text{Mass of Sodium}}{\text{Mass of } \text{Na}_2\text{CO}_3} = \frac{46}{106}$$

Now, to find the mass of sodium in $53 \, \text{g}$ of $\text{Na}_2\text{CO}_3$:

$$\text{Mass of Sodium}=\left(\frac{46}{106}\right)\times 53$$

Step 4: Perform the calculation:

$$\text{Mass of Sodium} = \frac{46 \times 53}{106} = \frac{2438}{106} \approx 23 \, \text{g}.$$

Final Answer:

The mass of sodium in $53 \, \text{g}$ of sodium carbonate is 23 g.

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Question 11:

What mass of chlorine is contained in $27.75 \, \text{g}$ of calcium chloride ($\text{CaCl}_2$)? Given that $\text{Ca} = 40 \, \text{g/mol}$, $\text{Cl} = 35.5 \, \text{g/mol}$.

Solution:

Step 1: Calculate the molar mass of  chloride (${\text{CaCl}}_2$)

The molar mass of $\text{CaCl}_2$ is given by:

$$\text{Molar Mass of } \text{CaCl}_2 = \text{Ca} + 2 \times \text{Cl}$$
$$\text{Molar Mass of } \text{CaCl}_2 = 40 + (2 \times 35.5) = 40 + 71 = 111 \, \text{g/mol}.$$

Step 2: Determine the mass of chlorine in one mole of CaCl2​

There are two chlorine ($\text{Cl}$) atoms in $\text{CaCl}_2$, and the mass of chlorine in one mole is:

$$\text{Mass of Chlorine}=2\times 35.5=71\text{g}\mathrm{.}$$

Step 3: Use the proportion to find the mass of chlorine in 27.75$$$\text{g}$ of CaCl$2_{}$​

If $111 \, \text{g}$ of $\text{CaCl}_2$ contains $71 \, \text{g}$ of chlorine, the mass of chlorine in $27.75\text{g is:}$

$$\text{Mass of Chlorine}=\left(\frac{71}{111}\right)\times 27.75$$

Step 4: Perform the calculation

1. Calculate the ratio:

   $$\frac{71}{111} \approx 0.6396$$
   

2. Multiply by $27.75$:

   $$0.6396\times 27.75\approx 17.75\text{g}\mathrm{.}$$

Final Answer:

The mass of chlorine in $27.75 \, \text{g}$ of calcium chloride is 17.75 g.

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Question 12:

Find the value of $x$ in $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ if $0.375 \, \text{mol}$ of the compound has a mass of $107.25 \, \text{g}$.

Solution:

Step 1: Write the formula for the molar mass of${\mathrm{ Na2}}_{}$${\text{CO}}_{}$$3_{}$$\cdot$$x$${\text{H}}_{}$$2_{}$$\text{O}$

The molar mass of $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ is given by:

$$M = M_{\text{Na}_2\text{CO}_3} + x \cdot M_{\text{H}_2\text{O}}$$

Where:

• $M_{\text{Na}_2\text{CO}_3} = 106 \, \text{g/mol}$ (calculated as $2 \times 23 + 12 + 3 \times 16$)
• $M_{\text{H}_2\text{O}} = 18 \, \text{g/mol}$
  

Step 2: Use the molar mass formula

From the question, the total mass of $0.375 \, \text{mol}$ is $107.25 \, \text{g}$. The molar mass of the compound can be calculated as:

$$M = \frac{\text{Mass}}{\text{Moles}}$$

Substitute the values:

$$M = \frac{107.25}{0.375} = 286 \, \text{g/mol}.$$

Thus:

$$286 = 106 + x \cdot 18$$

Step 3: Solve for x

Rearrange the equation to isolate $x$:

$$286 - 106 = x \cdot 18$$
$$180 = x \cdot 18$$

Divide through by 18:

$$x = \frac{180}{18} = 10$$

Final Answer:

The value of $x$ is 10.

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Question 13:

Calculate the number of molecules in $1.7 \, \text{g}$of ammonia ($\text{NH}_3$). Given $\text{N} = 14 \, \text{g/mol}, \text{H} = 1 \, \text{g/mol}$and Avogadro's constant $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$.

Solution:

Step 1: Calculate the molar mass of ammonia (

$\text{NH}_3$)

The molar mass of ammonia is:

$$M_{\text{NH}_3} = 14 + (3 \times 1) = 17 \, \text{g/mol}.$$

Step 2: Determine the number of moles of ammonia

The number of moles ($n$) is given by:

$$n = \frac{\text{Mass}}{\text{Molar Mass}}.$$

$$n = \frac{1.7}{17} = 0.1 \, \text{mol}.$$

Step 3: Calculate the number of molecules

The number of molecules is calculated using Avogadro's constant:

$$\text{Number of molecules} = n \times L.$$

Substitute the values:

$$\text{Number of molecules} = 0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}.$$

Final Answer:

The number of molecules in $1.7 \, \text{g}$ of ammonia is $6.02 \times 10^{22} \, \text{molecules}$.

---

Question 14:

Calculate the amount of substance (in moles) in:
(i) $7 \, \text{g}$ of nitrogen gas ($\text{N}_2$)
(ii) $6 \, \text{g}$ of ethanoic acid ($\text{CH}_3\text{COOH}$)

Solution:

Part (i): 7 g of nitrogen gas (${\text{N}}_{}$$2_{}$​)

Step 1: Calculate the molar mass of $$N2

The atomic mass of nitrogen ($\text{N}$) is $14 \, \text{g/mol}$. Since $\text{N}_2$ is diatomic:

$$\text{Molar Mass of } \text{N}_2 = 2 \times 14 = 28 \, \text{g/mol}.$$

Step 2: Calculate the number of moles

The number of moles is given by:

$$n=\frac{\text{Mass}}{\text{Molar Mass}}\mathrm{.}$$

Substitute the values:

$$n = \frac{7}{28} = 0.25 \, \text{mol}.$$

Part (ii): 6 g of ethanoic acid (CH3COOH)

Step 1: Calculate the molar mass of $\text{CH}_3\text{COOH}$

The molar mass of ethanoic acid is calculated as:

$$M_{\text{CH}_3\text{COOH}} = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \, \text{g/mol}.$$

Step 2: Calculate the number of moles

Substitute into the formula:

$$n = \frac{\text{Mass}}{\text{Molar Mass}}.$$

Substitute the values:

$$n = \frac{6}{60} = 0.1 \, \text{mol}.$$

Final Answer:

(i) The amount of substance in $7 \, \text{g}$ of nitrogen gas is $0.25\text{mol}\text{<br>}\text{.}\text{<br>}\text{<br>}$

(ii) The amount of substance in 6 g of ethanoic acid is 0.1 mol.

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Question 15:

Calculate the number of atoms in $18.5 \, \text{g}$ of magnesium metal. Given:

• Molar mass of magnesium ($\text{Mg}$) = $24.3 \, \text{g/mol}$
  
• Avogadro's number ($L$) = $6.0 \times 10^{23} \, \text{mol}^{-1}$
  

Solution:

Step 1: Calculate the number of moles of magnesium

The number of moles ($n$) is given by:

$$n = \frac{\text{Mass}}{\text{Molar Mass}}.$$

Substitute the values:

$$n = \frac{18.5}{24.3} \approx 0.761 \, \text{mol}.$$

Step 2: Calculate the number of atoms

The number of atoms is given by:

$$\text{Number of atoms}=n\times L\mathrm{..}$$

Substitute the values:

$$\text{Number of atoms} = 0.761 \times 6.0 \times 10^{23}.$$

Perform the calculation:

$$\text{Number of atoms} = 4.566 \times 10^{23}.$$

Final Answer:

The number of atoms in $18.5\text{g of magnesium is }$$4.57 \times 10^{23} \, \text{atoms}$.

.

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Question 16:

Calculate the number of chloride ions present in $65.0 \, \text{g}$ of calcium chloride ($\text{CaCl}_2$). Given:

• $\text{Ca}=40\text{g/mol}$
• $\text{Cl} = 35.5 \, \text{g/mol}$
  
• Avogadro's number ($L$) = $6.02 \times 10^{23} \, \text{particles/mol}$.

Solution:

Step 1: Calculate the molar mass of calcium chloride (

$\text{CaCl}_2$)

The molar mass of $\text{CaCl}_2$ is:

$$M_{\text{CaCl}_2} = M_{\text{Ca}} + 2 \times M_{\text{Cl}}$$ $$M_{\text{CaCl}_2} = 40 + (2 \times 35.5) = 40 + 71 = 111 \, \text{g/mol}.$$

Step 2: Calculate the number of moles of

$\text{CaCl}_2$

The number of moles ($n$) is given by:

$$n=\frac{\text{Mass}}{\text{Molar Mass}}\mathrm{.}$$

Substitute the values:

$$n = \frac{65.0}{111} \approx 0.586 \, \text{mol}.$$

Step 3: Determine the number of chloride ions

Each formula unit of $\text{CaCl}_2$ contains 2 chloride ions. Therefore, the total number of chloride ions is:

$$\text{Number of chloride ions} = n \times L \times 2.$$

Substitute the values:

$$\text{Number of chloride ions} = 0.586 \times 6.02 \times 10^{23} \times 2.$$

Perform the calculation:

$$\text{Number of chloride ions} = 7.06 \times 10^{23}.$$

Final Answer:

The number of chloride ions in $65.0 \, \text{g}$ of calcium chloride is $7.06\times 10^{23}\text{ions.}\text{<br>}\text{<br>}\text{<br>}$

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Question 17:

Calculate the number of chlorine atoms contained in $0.2 \, \text{g}$ of DDT ($\text{C}_{14}\text{H}_9\text{Cl}_5$).
Given:

• $\text{C} = 12 \, \text{g/mol}$
  
• $\text{H} = 1 \, \text{g/mol}$
  
• $\text{Cl} = 35.5 \, \text{g/mol}$
  
• Avogadro's number ($L$) = $6.02 \times 10^{23} \, \text{particles/mol}$.

Solution:

Step 1: Calculate the molar mass of $\text{C}_{14}\text{H}_9\text{Cl}_5$

The molar mass of DDT is given by:

$$M_{\text{C}_{14}\text{H}_9\text{Cl}_5} = (14 \times 12) + (9 \times 1) + (5 \times 35.5)$$
$$M_{\text{C}_{14}\text{H}_9\text{Cl}_5} = 168 + 9 + 177.5 = 354.5 \, \text{g/mol}.$$

Step 2: Calculate the number of moles of DDT

The number of moles ($n$) is:

$$n = \frac{\text{Mass}}{\text{Molar Mass}}$$

Substitute the values:

$$n = \frac{0.2}{354.5} \approx 5.64 \times 10^{-4} \, \text{mol}.$$

Step 3: Determine the number of chlorine atoms

Each molecule of $\text{C}_{14}\text{H}_9\text{Cl}_5$ contains 5 chlorine atoms. Therefore, the total number of chlorine atoms is:

$$\text{Number of chlorine atoms} = n \times L \times 5$$

Substitute the values:

$$\text{Number of chlorine atoms} = 5.64 \times 10^{-4} \times 6.02 \times 10^{23} \times 5$$

Perform the calculations step by step:

1. $5.64 \times 10^{-4} \times 6.02 \times 10^{23} = 3.397 \times 10^{20}$
   
2. $3.397 \times 10^{20} \times 5 = 1.698 \times 10^{21}$
   

Final Answer:

The number of chlorine atoms in $0.2 \, \text{g}$ of DDT is $1.70 \times 10^{21} \, \text{atoms}$.

---

Question 18:

Calculate the number of hydrogen atoms in $3.7 \, \text{g}$ of diethyl ether vapor ($(\text{C}_2\text{H}_5)_2\text{O}$.
Given:

• $\text{C} = 12 \, \text{g/mol}$
  
• $\text{H} = 1 \, \text{g/mol}$
  
• $\text{O} = 16 \, \text{g/mol}$
  
• Avogadro's number ($L$) = $6.02 \times 10^{23} \, \text{particles/mol}$

Solution:

Step 1: Determine the molecular formula and molar mass of 

$(\text{C}_2\text{H}_5)_2\text{O}$

Diethyl ether consists of two ethyl groups ($\text{C}_2\text{H}_5$) and one oxygen atom. Its molecular formula is $\text{C}_4\text{H}_{10}\text{O}$.

The molar mass is calculated as:

$$M_{({\text{C}}_2{\text{H}}_5{)}_2\text{O}}=(4\times 12)+(10\times 1)+(1\times 16)=48+10+16=74\text{g/mol}\mathrm{.}$$

Step 2: Calculate the number of moles of diethyl ether

The number of moles ($n$) is given by:

$$n=\frac{\text{Mass}}{\text{Molar Mass}}\mathrm{.}$$

Substitute the values:

$$n=\frac{3.7}{74}\approx 0.05\text{mol}\mathrm{.}$$

Step 3: Determine the number of hydrogen atoms in one molecule of diethyl ether

Each molecule of diethyl ether ($\text{C}_4\text{H}_{10}\text{O}$) contains 10 hydrogen atoms.

Step 4: Calculate the number of hydrogen atoms

The number of hydrogen atoms is given by:

$$\text{Number of hydrogen atoms} = n \times L \times \text{Number of H atoms per molecule}.$$

Substitute the values:

$$\text{Number of hydrogen atoms} = 0.05 \times 6.02 \times 10^{23} \times 10.$$

Perform the calculations step by step:

1. $0.05 \times 6.02 \times 10^{23} = 3.01 \times 10^{22}$
   
2. $3.01 \times 10^{22} \times 10 = 3.01 \times 10^{23}$.

Final Answer:

The number of hydrogen atoms in $3.7 \, \text{g}$ of diethyl ether vapor is $3.01 \times 10^{23} \, \text{atoms}$.

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Molar Volume and Gases: The Special Case for Gaseous Substances

Molar volume is defined as the volume occupied by one mole of a gas at a given temperature and pressure. At standard temperature and pressure (STP), which is defined as:

• Temperature: $273.15 \, \text{K}$ (0°C)
• Pressure: $1 \, \text{atm}$ or $101.325 \, \text{kPa}$,

the molar volume of an ideal gas is approximately:

$$\text{Molar Volume (STP)} = 22.4 \, \text{L/mol}.$$

At standard ambient temperature and pressure (SATP), defined as:

• Temperature: $298.15 \, \text{K}$ (25°C),
• Pressure: $1 \, \text{atm}$,

the molar volume is approximately $24.0 \, \text{L/mol}$.

Formula

The relationship between the molar volume, number of moles ($n$), and volume ($V$) of a gas can be expressed as:

$$V = n \times \text{Molar Volume}.$$

Derivation from Ideal Gas Law

The molar volume of a gas can also be derived from the ideal gas law:

$$PV=nRT,$$

where:

• $P$ = pressure ($\text{atm),}$
• $V$ = volume ($\text{L}$),
• $n$ = number of moles,
• $R$ = ideal gas constant ($0.0821 \, \text{L·atm·mol}^{-1}\text{K}^{-1}$),
• $T$ = temperature ($\text{K}$).

At STP ($P = 1 \, \text{atm}$, $T=273.15\text{K), one mole of gas gives:}$

$$V = \frac{nRT}{P}.$$

Substituting the values for $n = 1 \, \text{mol}$:

$$V = \frac{(1)(0.0821)(273.15)}{1} = 22.4 \, \text{L/mol}.$$

Applications

• Stoichiometry in Chemical Reactions:

The molar volume allows the direct conversion between moles of gas and volume in liters under standard conditions.

Example:

For the reaction $\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$,

one mole of $\text{H}_2$ reacts with one mole of $\text{Cl}_2$ to produce two moles of $\text{HCl}$. Using molar volume, this means:

• $22.4\text{L of }$$\text{H}_2$ reacts with $22.4 \, \text{L}$ of $\text{Cl}_2$ to produce $44.8\text{L of }$$\text{HCl}$ at STP.

• Gas Density Calculations:

The molar volume is useful for determining the density ($\rho$) of a gas:

$$\rho =\frac{\text{Molar Mass}}{\text{Molar Volume}}\mathrm{.}$$

• Volumetric Analysis:

Molar volume simplifies calculations in experiments involving gas collection and analysis.

---

Sample Questions and Answers on Molar Volume:

Question 1

Calculate the amount of $\text{CO}_2$ gas in $28 \, \text{dm}^3$ at STP.
Given:

• $\text{Molar Volume} (\text{Vm}) = 22.4 \, \text{dm}^3/\text{mol}$

Solution:

The number of moles ($n$) of a gas can be calculated using the formula:

$$n = \frac{\text{Volume}}{\text{Molar Volume}}.$$

Substitute the given values:

$$n = \frac{28}{22.4}.$$

Perform the division:

$$n = 1.25 \, \text{mol}.$$

Final Answer:

The amount of $\text{CO}_2$ in $28 \, \text{dm}^3$ of gas at STP is 1.25 moles.

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Question 2

Determine the relative molecular mass of trichloromethane ($\text{CHCl}_3$) given:

• Mass of trichloromethane = $0.12 \, \text{g}$
  
• Volume of vapor formed = $22.4 \, \text{cm}^3$
  
• Molar volume at STP = $22.4 \, \text{dm}^3$ = $22,400 \, \text{cm}^3$.

Solution

Step 1: Calculate the number of moles of vapor

The number of moles ($n$) is given by the formula:

$$n=\frac{\text{Volume of gas}}{\text{Molar Volume}}\mathrm{.}$$

Substitute the given values:

$$n = \frac{22.4}{22,400}.$$

Simplify:

$$n = 0.001 \, \text{mol}.$$

Step 2: Calculate the molar mass

The molar mass ($M$) is determined using:

$$M = \frac{\text{Mass}}{\text{Number of moles}}.$$

Substitute the values:

$$M = \frac{0.12}{0.001}.$$

Simplify:

$$M = 120 \, \text{g/mol}.$$

Final Answer

The relative molecular mass of trichloromethane ($\text{CHCl}_3$) is 120 g/mol.

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Question 3:

A container holds $3.7 \, \text{g}$ of diethyl ether vapor ($(\text{C}_2\text{H}_5)_2\text{O}$). Calculate:
(a) The number of moles of $(\text{C}_2\text{H}_5)_2\text{O}$.
(b) The volume of the vapor at STP.
(c) The number of molecules of $(\text{C}_2\text{H}_5)_2\text{O}$.

Data provided:

• Avogadro's number ($L$) = $6.02\times 10^{23}{\text{mol}}^{-1}$.
• Molar volume ($V_m$) at STP = $22.4{\text{dm}}^3\mathrm{/}\text{mol}$.

Solution

• $\text{Molar Mass of }({\text{C}}_2{\text{H}}_5{)}_2\text{O}=(4\times 12)+(10\times 1)+(1\times 16)=74\text{g/mol}$

(a) Calculate the number of moles of  (${\text{C}}_{}$$2_{}$${\text{H}}_{}$$5_{}$${)}_{}$$2_{}$$\text{O}$:

The number of moles ($n$) is given by:

$$n=\frac{\text{Mass}}{\text{Molar Mass}}\mathrm{.}$$

Substitute the values:

$$n = \frac{3.7}{74}.$$

Perform the division:

$$n = 0.05 \, \text{mol}.$$

(b) Calculate the volume of the vapor at STP:

The volume of the vapor is given by:

$$V = n \times V_m.$$

Substitute the values:

$$V = 0.05 \times 22.4.$$

Simplify:

$$V = 1.12 \, \text{dm}^3.$$

(c) Calculate the number of molecules of $(\text{C}_2\text{H}_5)_2\text{O}$:

The number of molecules is given by:

$$\text{Number of molecules} = n \times L.$$

Substitute the values:

$$\text{Number of molecules} = 0.05 \times 6.02 \times 10^{23}.$$

Simplify:

$$\text{Number of molecules} = 3.01 \times 10^{22}.$$

Final Answers

(a) The number of moles of $(\text{C}_2\text{H}_5)_2\text{O}$ is 0.05 mol.
(b) The volume of the vapor at STP is 1.12 dm³.
(c) The number of molecules of $(\text{C}_2\text{H}_5)_2\text{O}$ is $3.01 \times 10^{22}$ molecules.

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Question 4:

Calculate the molar mass of a gaseous hydrocarbon weighing 11 g, which occupies $5.6 \, \text{dm}^3$ at STP.
Given:

• Volume ($V$) = $5.6 \, \text{dm}^3$
  
• Molar Volume ($V_m$) = $22.4{\text{dm}}^3\mathrm{/}\text{mol}$
• Mass = $11\text{g}$

Solution:

Step 1: Calculate the number of moles of the gas

The number of moles ($n$) is calculated using:

$$n = \frac{\text{Volume}}{\text{Molar Volume}}.$$

Substitute the values:

$$n = \frac{5.6}{22.4}.$$

Simplify:

$$n = 0.25 \, \text{mol}.$$

Step 2: Calculate the molar mass

The molar mass ($M$) is given by:

$$M = \frac{\text{Mass}}{\text{Number of moles}}.$$

Substitute the values:

$$M = \frac{11}{0.25}.$$

Simplify:

$$M = 44 \, \text{g/mol}.$$

Final Answer:

The molar mass of the gaseous hydrocarbon is 44 g/mol.

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Question 5:

Calculate the molar mass and atomic mass of $X$ in $17.75 \, \text{g}$ of a gaseous compound $X_2O_5$, which occupies $2.8 \, \text{dm}^3$ at STP.
Given:

• $\text{Volume (V)} = 2.8 \, \text{dm}^3$
  
• $\text{Molar Volume (Vm)} = 22.4 \, \text{dm}^3/\text{mol}$
  
• $\text{Mass} = 17.75 \, \text{g}$
  
• $\text{Oxygen Atomic Mass (O)} = 16$.

Solution:

Step 1: Calculate the number of moles of $X_2O_5$:

The number of moles ($n$) is given by:

$$n = \frac{\text{Volume}}{\text{Molar Volume}}.$$

Substitute the values:

$$n = \frac{2.8}{22.4}.$$

Simplify:

$$n = 0.125 \, \text{mol}.$$

Step 2: Calculate the molar mass of $X_2O_5$:

The molar mass ($M$) is given by:

$$M = \frac{\text{Mass}}{\text{Number of moles}}.$$

Substitute the values:

$$M = \frac{17.75}{0.125}.$$

Simplify:

$$M = 142 \, \text{g/mol}.$$

Step 3: Determine the atomic mass of $\mathrm{X:}$

The molar mass of $X_2O_5$ can also be written as:

$$M=(2\times \text{Atomic Mass of }X)+(5\times \text{Atomic Mass of }O)\mathrm{.}$$

Substitute the known values:

$$142 = (2 \times X) + (5 \times 16).$$

Simplify:

$$142 = (2 \times X) + 80.$$

Solve for $X$:

$$2X = 142 - 80 = 62.$$
$$X = \frac{62}{2} = 31.$$

Final Answer:

• The molar mass of $X_2O_5$ is 142 g/mol.
• The atomic mass of $X$ is 31.

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Question 6:

A gas $\text{X}_2\text{O}_3$ occupies $5.6 \, \text{dm}^3$ at STP. Calculate:

1. The molar mass of $\text{X}_2\text{O}_3$.
2. The atomic mass of $\text{X}$, given that the mass of the gas is $16 \, \text{g}$.

Data provided:

• Volume ($V$) = $5.6 \, \text{dm}^3$
  
• Molar Volume ($V_m$) = $22.4 \, \text{dm}^3/\text{mol}$
  
• Mass = $16 \, \text{g}$
  
• Oxygen atomic mass ($O$) = $16 \, \text{g/mol}$.

Solution

Step 1: Calculate the number of moles of ${\text{X}}_2{\text{O}}_3$

The number of moles ($n$) is given by:

$$n = \frac{\text{Volume}}{\text{Molar Volume}}.$$

Substitute the values:

$$n = \frac{5.6}{22.4}.$$

Simplify:

$$n = 0.25 \, \text{mol}.$$

Step 2: Calculate the molar mass of ${\text{X}}_2{\text{O}}_3$

The molar mass ($M$) is given by:

$$M = \frac{\text{Mass}}{\text{Number of moles}}.$$

Substitute the values:

$$M = \frac{16}{0.25}.$$

Simplify:

$$M = 64 \, \text{g/mol}.$$

Step 3: Determine the atomic mass of X:

The molar mass of $\text{X}_2\text{O}_3$ can also be expressed as:

$$M = (2 \times \text{Atomic Mass of } X) + (3 \times \text{Atomic Mass of } O).$$

Substitute the known values:

$$64 = (2 \times X) + (3 \times 16).$$

Simplify:

$$64 = (2 \times X) + 48.$$

Solve for $X$:

$$2X = 64 - 48 = 16.$$
$$X = \frac{16}{2} = 8.$$

Final Answer

1. The molar mass of $\text{X}_2\text{O}_3$ is 64 g/mol.
2. The atomic mass of $\text{X}$ is 8.

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Question 7:

A gaseous hydrocarbon, $\text{C}_x\text{H}_8$, has a mass of $7 \, \text{g}$ and occupies a volume of $2.8 \, \text{dm}^3$ at STP. Calculate:

1. The relative molar mass ($M$) of $\text{C}_x\text{H}_8$.
2. The value of $x$ in $\text{C}_x\text{H}_8$.

Data Provided:

• Volume ($V$) = $2.8 \, \text{dm}^3$
  
• Molar Volume ($V_m$) = $22.4 \, \text{dm}^3/\text{mol}$
  
• Mass = $7\text{g}$
• Carbon atomic mass ($\text{C}$) = $12$
  
• Hydrogen atomic mass ($\text{H}$) = $1$.

Solution

Step 1: Calculate the number of moles of $\text{C}_x\text{H}_8$:

The number of moles ($n$) is given by:

$$n = \frac{\text{Volume}}{\text{Molar Volume}}.$$

Substitute the values:

$$n = \frac{2.8}{22.4}.$$

Simplify:

$$n = 0.125 \, \text{mol}.$$

Step 2: Calculate the relative molar mass (M) of $\text{C}_x\text{H}_8$:

The molar mass ($M$) is given by:

$$M = \frac{\text{Mass}}{\text{Number of moles}}.$$

Substitute the values:

$$M = \frac{7}{0.125}.$$

Simplify:

$$M = 56 \, \text{g/mol}.$$

Step 3: Determine the value of x

The molecular formula of $\text{C}_x\text{H}_8$ is expressed as:

$$M = (x \times \text{Atomic Mass of C}) + (8 \times \text{Atomic Mass of H}).$$

Substitute the known values:

$$56 = (x \times 12) + (8 \times 1).$$

Simplify:

$$56 = 12x + 8.$$

Solve for $x$:

$$12x = 56 - 8 = 48.$$
$$x = \frac{48}{12} = 4.$$

Final Answer

1. The relative molar mass of $\text{C}_x\text{H}_8$ is 56 g/mol.
2. The value of $x$ is 4, so the molecular formula is $\text{C}_4\text{H}_8$.

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Question 8:

A vapour with molecular formula $(C_2H_5)_xO$ has a mass of $3.7 \, \text{g}$ and occupies a volume of $1.12 \, \text{dm}^3$ at STP. Calculate:

1. The relative molar mass ($M$) of $(C_2{H}_5{)}_x\mathrm{O.}$
2. The value of $x$ in $(C_2H_5)_xO$.

Data Provided:

• Volume ($V$) = $1.12 \, \text{dm}^3$
  
• Molar Volume ($V_m$) = $22.4 \, \text{dm}^3/\text{mol}$
  
• Mass = $3.7 \, \text{g}$
  
• Atomic masses: $\text{C}=\mathrm{12, }$$\text{H} = 1$, $\text{O} = 16$.

Solution

Step 1: Calculate the number of moles of $(C_2H_5)_xO$:

The number of moles ($n$) is given by:

$$n = \frac{\text{Volume}}{\text{Molar Volume}}.$$

Substitute the values:

$$n = \frac{1.12}{22.4}.$$

Simplify:

$$n = 0.05 \, \text{mol}.$$

Step 2: Calculate the relative molar mass ($M$) of $(C_2H_5)_xO$:

The molar mass ($M$) is given by:

$$M = \frac{\text{Mass}}{\text{Number of moles}}.$$

Substitute the values:

$$M = \frac{3.7}{0.05}.$$

Simplify:

$$M = 74 \, \text{g/mol}.$$

Step 3: Determine the value of $x$:

The molecular formula $(C_2H_5)_xO$ can be expressed as:

$$M = (x \times \text{Molar Mass of } C_2H_5) + (\text{Molar Mass of } O).$$

The molar mass of $C_2H_5$ is:

$$(2 \times 12) + (5 \times 1) = 24 + 5 = 29.$$

Substitute the values:

$$74 = (x \times 29) + 16.$$

Simplify:

$$74 = 29x + 16.$$

Solve for $x$:

$$29x = 74 - 16 = 58.$$
$$x = \frac{58}{29} = 2.$$

Final Answer

1. The relative molar mass of $(C_2H_5)_xO$ is 74 g/mol.
2. The value of $x$ is 2, so the molecular formula is $(C_2H_5)_2O$.

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Faraday Constant (F)

Introduction

The Faraday constant is a fundamental concept in electrochemistry, representing the total charge of one mole of electrons. It is widely used in calculations involving electrolysis, redox reactions, and other electrochemical processes.

Definition of the Faraday Constant

The Faraday constant ($F$) is the magnitude of electric charge carried by one mole of electrons. Mathematically, it is given by:

$$F = N_A \cdot e$$

Where:

• $N_A$ = Avogadro's number ($6.022\times 10^{23}{\text{mol}}^{-\mathrm{1)}}$
• $e$ = Charge of a single electron ($1.602 \times 10^{-19} \, \text{C}$)

Substituting these values:

$$F = 6.022 \times 10^{23} \cdot 1.602 \times 10^{-19} = 96485 \, \text{C/mol}.$$

Thus, the Faraday constant is approximately $96485 \, \text{C/mol}$ (coulombs per mole of electrons).

Applications of the Faraday Constant

Electrolysis
In electrolysis, the Faraday constant is used to relate the amount of charge passed through an electrolyte to the amount of substance deposited or dissolved at an electrode.

$$Q = nF$$

Where $Q$ is the total charge, $n$ is the number of moles of electrons, and $F$ is the Faraday constant.

Faraday's Laws of Electrolysis

• First Law: The amount of substance deposited at an electrode is directly proportional to the charge passed.
• Second Law: For different substances, the amounts of substances deposited are proportional to their equivalent weights.

Electrochemical Cell Calculations
The Faraday constant is essential in determining the standard Gibbs free energy ($\Delta G^\circ$) of electrochemical reactions:

$$\Delta G^\circ = -nFE^\circ$$

Where $E^\circ$ is the standard electrode potential.

Batteries and Energy Storage
The constant helps in quantifying the energy capacity and efficiency of electrochemical cells.

Significance in Modern Chemistry

The Faraday constant provides a bridge between the macroscopic world of measurable quantities (such as charge and mass) and the microscopic world of individual particles like electrons. It is indispensable in advancing our understanding of chemical and physical processes involving charge transfer.

Calculation involving Faraday Constant (F)

Question 1: 

Calculate the total charge in coulombs carried by $3 \, \text{mol}$ of electrons.
Given:

• Faraday constant ($F$) = $96485 \, \text{C/mol}$.

Solution:

The total charge ($Q$) is calculated using the formula:

$$Q = n \times F,$$

where:

• $n$ = number of moles of electrons,
• $F$ = Faraday constant.

Substitute the given values:

$$Q = 3 \times 96485.$$

Simplify:

$$Q = 289455 \, \text{C}.$$

Final Answer:

The total charge carried by $3 \, \text{mol}$ of electrons is 289455 coulombs.

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Question 2:

Calculate the time needed to pass $0.5 \, \text{mol}$ of electrons using a current of $25 \, \text{A}$.
Given:

• Number of moles of electrons ($n$) = $0.5 \, \text{mol}$,
• Current ($I$) = $25 \, \text{A}$,
• Faraday constant ($F$) = $96485 \, \text{C/mol}$.

Solution:

Step 1: Calculate the total charge ($Q$) needed:

The total charge ($Q$) is given by:

$$Q=n\times F,$$

where:

• $n = 0.5 \, \text{mol}$,
• $F = 96485 \, \text{C/mol}$.

Substitute the values:

$$Q = 0.5 \times 96485 = 48242.5 \, \text{C}.$$

Step 2: Relate charge to time using the formula:

The relationship between charge, current, and time is:

$$Q=I\times t,$$

where:

• $Q$ = charge in coulombs,
• $I$ = current in amperes,
• $t$ = time in seconds.

Rearrange for $t$:

$$t = \frac{Q}{I}.$$

Substitute the known values:

$$t = \frac{48242.5}{25}.$$

Simplify:

$$t = 1929.7 \, \text{seconds}.$$

Step 3: Convert time to minutes (optional):

To express the time in minutes:

$$t = \frac{1929.7}{60} \approx 32.16 \, \text{minutes}.$$

Final Answer:

The time needed to pass $0.5\text{mol of electrons using a current of }$$25 \, \text{A}$ is approximately 1929.7 seconds (or 32.16 minutes).

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Question 3:

Calculate the amount of electrons in moles needed to carry a total charge of 48,250 coulombs.
Given:

• Total charge ($Q$) = 48,250 C,
• Faraday constant ($F$) = 96,485 C/mol (the charge carried by 1 mole of electrons).

Solution:

The number of moles of electrons ($n$) can be calculated using the formula:

$$n = \frac{Q}{F}$$

where:

• $Q$ = total charge in coulombs,
• $F$ = Faraday constant ($96,485 \, \text{C/mol}$).

Substitute the values:

$$n = \frac{48,250}{96,485}.$$

Simplify:

$$n \approx 0.5 \, \text{mol}.$$

Final Answer:

The amount of electrons in moles needed to carry a total charge of 48,250 coulombs is approximately 0.5 mol.

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Question 4:

How many Faradays of electricity are carried by $1.505 \times 10^{23}$ electrons?
Given:

• Avogadro's number ($N_A$) = $6.022 \times 10^{23} \, \text{mol}^{-1}$,
• Faraday constant ($F$) = $96,485 \, \text{C/mol}$,

Solution:

Step 1: Calculate the number of moles of electrons.

The number of moles of electrons ($n$) is calculated using the relationship between the number of electrons and Avogadro's number:

$$n=\frac{\text{Number of electrons}}{N_A}\mathrm{.}$$

Substitute the given values:

$$n = \frac{1.505 \times 10^{23}}{6.022 \times 10^{23}} = 0.25 \, \text{mol}.$$

Step 2: Calculate the number of Faradays.

One mole of electrons carries 1 Faraday of charge, so the number of Faradays is equal to the number of moles of electrons.

Thus, the number of Faradays of electricity carried by $1.505 \times 10^{23}$ electrons is:

$$\text{Faradays} = 0.25 \, \text{mol}.$$

Final Answer:

The number of Faradays of electricity carried by $1.505 \times 10^{23}$ electrons is 0.25 Faradays.

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Question 5:

How many electrons carry a total charge of 48,250 C?
Given:

• Charge of a single electron ($e$) = $1.602 \times 10^{-19} \, \text{C}$,
• Total charge ($Q$) = 48,250 C.

Solution:

Step 1: Calculate the number of electrons.

The number of electrons ($n$) is given by:

$$n=\frac{Q}{\mathrm{e}}$$

Where:

• $Q$ = total charge in coulombs,
• $e$ = charge of a single electron.

Substitute the given values:

$$n=\frac{48,250}{1.602\times 10^{-19}}\mathrm{.}$$

Simplify:

$$n = 3.01 \times 10^{23} \, \text{electrons}.$$

Final Answer:

The number of electrons that carry a total charge of 48,250 C is approximately $3.01 \times 10^{23}$electrons.

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Question 6:

Given that $0.25 \, \text{mol}$ of electrons pass through an electrical wire using a current of $0.25 \, \text{A}$, calculate:
(i) the total charge in coulombs carried by the electrons,
(ii) the time taken for the electrons to pass through.

Solution:

Part (i): Total charge in coulombs carried by the electrons.

The total charge ($Q$) carried by $n$ moles of electrons is given by:

$$Q=n\times F,$$

where:

• $n = 0.25 \, \text{mol}$,
• $F=96,485\text{C/mol (Faraday constant).}$

Substitute the values:

$$Q = 0.25 \times 96,485 = 24,121.25 \, \text{C}.$$

So, the total charge carried by $0.25 \, \text{mol}$ of electrons is 24,121.25 coulombs.

Part (ii): Time taken for the electrons to pass through.

The relationship between charge, current, and time is:

$$Q=I\times t,$$

where:

• $Q$ = charge in coulombs,
• $I$ = current in amperes,
• $t$ = time in seconds.

Rearrange for $t$:

$$t=\frac{Q}{I}\mathrm{.}$$

Substitute the values:

$$t = \frac{24,121.25}{0.25} = 96,485 \, \text{seconds}.$$

To convert the time into hours:

$$t = \frac{96,485}{3600} \approx 26.8 \, \text{hours}.$$

Final Answer:

(i) The total charge carried by $0.25 \, \text{mol}$ of electrons is 24,121.25 coulombs.
(ii) The time taken for the electrons to pass through is approximately 96,485 seconds (or 26.8 hours).

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Amount of Concentration or Molarity

Introduction to Molarity
Molarity, also known as the amount of concentration, is one of the most commonly used methods to express the concentration of a solution in chemistry. It provides a measure of the amount of solute (in moles) dissolved in a specific volume of solution. This unit is particularly useful for quantitative analysis in chemical reactions and laboratory experiments.

Definition of Molarity

Molarity ($M$) is defined as the number of moles of solute ($n$) dissolved in one liter ($L$) of solution. It is mathematically expressed as:

$$M=\frac{n}{V},$$

where:

• $M$ = Molarity (mol/L),
• $n$ = Number of moles of solute (mol),
• $V$ = Volume of the solution (L).

Key Components of Molarity

1. Solute: The substance being dissolved (e.g., salt, sugar, acid).
2. Solvent: The liquid in which the solute is dissolved (commonly water).
3. Solution: The homogeneous mixture formed when solute is dissolved in solvent.

Units of Molarity

Molarity is expressed in moles per liter (mol/L), commonly abbreviated as $M$. For example, a 1 M solution contains 1 mole of solute per liter of solution.

How to Calculate Molarity

1. Determine the number of moles of solute:
   Use the molar mass of the solute to convert from grams to moles.

   $$n = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}}.$$
   

2. Measure the volume of the solution:
   Ensure the volume is in liters (1 L = 1000 mL).

3. Apply the molarity formula:
   Substitute the values of $n$ and $V$ into the formula $M = \frac{n}{V}$.

Examples of Molarity Calculation

1. Example 1: What is the molarity of a solution containing 0.5 moles of NaCl dissolved in 1 liter of solution?

   $$M = \frac{n}{V} = \frac{0.5}{1} = 0.5 \, \text{M}.$$
   

2. Example 2: Calculate the molarity of a solution prepared by dissolving 20 g of NaOH (molar mass = 40 g/mol) in 500 mL of solution.

   • Number of moles ($n$): $$n = \frac{\text{mass}}{\text{molar mass}} = \frac{20}{40} = 0.5 \, \text{mol}.$$
     
   • Volume ($V$) in liters: $$V = \frac{500}{1000} = 0.5 \, \text{L}.$$
     
   • Molarity ($M$): $$M = \frac{n}{V} = \frac{0.5}{0.5} = 1 \, \text{M}.$$
     

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Sample Questions and Answers on Molarity

Question 1:

Find the molarity of a solution in which 0.5 moles of NaOH are dissolved in 100 cm³ of solution.

Solution:

To calculate molarity ($M$), use the formula:

$$M=\frac{n}{V}<br>$$

where:

• $n$ = Number of moles of solute (mol),
• $V$ = Volume of solution (L).

Step 1: Convert volume to liters

The given volume is 100 cm³. Convert this to liters:

$$V = \frac{100}{1000} = 0.1 \, \text{L}.$$

Step 2: Apply the formula

Substitute the values into the formula:

$$M = \frac{n}{V} = \frac{0.5}{0.1} = 5 \, \text{M}.$$

Final Answer:

The molarity of the NaOH solution is 5 M.

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Question 2:

Calculate the concentration in moles per dm³ of 8 g of NaOH dissolved in 500 cm³ of solution.
$\text{Na} = 23, \, \text{O} = 16, \, \text{H} = 1$

Solution:

Step 1: Calculate the molar mass of NaOH

$$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}.$$

Step 2: Calculate the number of moles of NaOH

Use the formula:

$$n=\frac{m}{M}<br>$$

where:

• $n$ = Number of moles (mol),
• $m$ = Mass of solute (g),
• $M$ = Molar mass of solute (g/mol).

Substitute the values:

$$n = \frac{8}{40} = 0.2 \, \text{mol}.$$

Step 3: Convert the volume to dm³

The given volume is 500 cm³. Convert this to dm³:

$$V=\frac{500}{1000}=0.5\text{dm³}\mathrm{.}\mathrm{<br>}$$

Step 4: Calculate the concentration

Use the formula:

$$C = \frac{n}{V},$$

where:

• $C$ = Concentration (mol/dm³),
• $n$ = Number of moles (mol),
• $V$ = Volume of solution (dm³).

Substitute the values:

$$C = \frac{0.2}{0.5} = 0.4 \, \text{mol/dm³}.$$

Final Answer:

The concentration of the NaOH solution is 0.4 mol/dm³.

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Question 3:

Calculate the mass of sodium chloride in 200 cm³ of a $0.5 \, \text{mol/dm}^3$ NaCl solution.

$$\text{Na} = 23, \, \text{Cl} = 35.5$$

Solution:

Step 1: Molar mass of NaCl

The molar mass of sodium chloride is:

$$\text{Molar mass of NaCl} = 23 + 35.5 = 58.5 \, \text{g/mol}.$$

Step 2: Volume of solution in dm³

Convert 200 cm³ to dm³:

$$V=\frac{200}{1000}=0.2\text{dm³}\mathrm{.}$$

Step 3: Calculate the number of moles of NaCl

Using the formula:

$$n=C\times V,$$

where:

• $n$ = Number of moles (mol),
• $C$ = Concentration of the solution ($\text{mol/dm}^3$),
• $V$ = Volume of the solution ($\text{dm}^3$).

Substitute the values:

$$n=0.5\times 0.2=0.1\text{mol}\mathrm{.}$$

Step 4: Calculate the mass of NaCl

Using the formula:

$$m=n\times M,$$

where:

• $m$ = Mass of solute (g),
• $n$ = Number of moles (mol),
• $M$ = Molar mass (g/mol).

Substitute the values:

$$m=0.1\times 58.5=5.85\text{g}\mathrm{.}$$

Final Answer:

The mass of sodium chloride in the solution is 5.85 g.

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Question 4:

How many grams of alkali (NaOH) are present in 150 cm³ of a $3 \, \text{mol/dm}^3$ NaOH solution?

Solution:

Step 1: Molar mass of NaOH

The molar mass of sodium hydroxide is:

$$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}.$$

Step 2: Convert the volume to dm³

Convert 150 cm³ to dm³:

$$V = \frac{150}{1000} = 0.15 \, \text{dm³}.$$

Step 3: Calculate the number of moles of NaOH

Using the formula:

$$n=C\times V,$$

where:

• $n$ = Number of moles (mol),
• $C$ = Concentration of the solution ($\text{mol/dm}^3$),
• $V$ = Volume of the solution ($\text{dm}^3$).

Substitute the values:

$$n = 3 \times 0.15 = 0.45 \, \text{mol}.$$

Step 4: Calculate the mass of NaOH

Using the formula:

$$m=n\times M,$$

where:

• $m$ = Mass of solute (g),
• $n$ = Number of moles (mol),
• $M$ = Molar mass (g/mol).

Substitute the values:

$$m=0.45\times 40=18\text{g}\mathrm{.}$$

Final Answer:

The mass of alkali (NaOH) present in the solution is 18 g.

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Question 5:

What volume of $0.300 \, \text{mol/dm}^3$ $\text{H}_2\text{SO}_4$ would contain 1.5 g of $\text{H}_2\text{SO}_4$?
$\text{H} = 1, \, \text{S} = 32, \, \text{O} = 16$

Solution:

Step 1: Calculate the molar mass of H2​SO4​

$$\text{Molar mass of } \text{H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \, \text{g/mol}.$$

Step 2: Calculate the number of moles of H2​SO4​

Using the formula:

$$n=\frac{m}{M}<br>$$

where:

• $n$ = Number of moles (mol),
• $m$ = Mass of $\text{H}_2\text{SO}_4$ (g),
• $M$ = Molar mass (g/mol).

Substitute the values:

$$n=\frac{1.5}{98}\approx 0.0153\text{mol}\mathrm{.}$$

Step 3: Calculate the volume of solution

Using the formula:

$$V=\frac{n}{C}<br>$$

where:

• $V$ = Volume of solution ($\text{dm}^3$),
• $n$ = Number of moles (mol),
• $C$ = Concentration ($\text{mol/dm}^3$).

Substitute the values:

$$V=\frac{0.0153}{0.300}=0.051{\text{dm}}^3\mathrm{.}$$

Convert to cm³:

$$V = 0.051 \times 1000 = 51 \, \text{cm}^3.$$

Final Answer:

The volume of $0.300 \, \text{mol/dm}^3$ $\text{H}_2\text{SO}_4$ solution required is 51 cm³.

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Question 6:

What volume of a $0.2 \, \text{mol/dm}^3$ solution of sodium hydroxide will, on evaporation, yield $5.0 \, \text{g}$ of solid NaOH?

$$\text{Na} = 23, \, \text{O} = 16, \, \text{H} = 1$$

Solution:

Step 1: Calculate the molar mass of NaOH

$$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}.$$

Step 2: Calculate the number of moles of NaOH

Using the formula:

$$n=\frac{m}{M}<br>$$

where:

• $n$ = Number of moles (mol),
• $m$ = Mass of NaOH (g),
• $M$ = Molar mass (g/mol).

Substitute the values:

$$n = \frac{5.0}{40} = 0.125 \, \text{mol}.$$

Step 3: Calculate the volume of solution

Using the formula:

$$V=\frac{n}{C}<br>$$

where:

• $V$ = Volume of solution ($\text{dm}^3$),
• $n$ = Number of moles (mol),
• $C$ = Concentration of solution ($\text{mol/dm}^3$).

Substitute the values:

$$V = \frac{0.125}{0.2} = 0.625 \, \text{dm}^3.$$

Convert to cm³:

$$V=0.625\times 1000=625{\text{cm}}^3\mathrm{.}$$

Final Answer:

The volume of $0.2 \, \text{mol/dm}^3$ sodium hydroxide solution required is 625 cm³.

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Question 7:

100 cm³ of $0.1 \, \text{mol/dm}^3$ NaOH is mixed with 100 cm³ of $0.15 \, \text{mol/dm}^3$ NaOH. Calculate the concentration of the resulting solution.

Solution:

Step 1: Calculate the moles of NaOH in each solution

Using the formula:

$$n=C\times V,$$

where:

• $n$ = Number of moles (mol),
• $C$ = Concentration ($\text{mol/dm}^3$),
• $V$ = Volume ($\text{dm}^3$).

For the first solution:

$$n_1 = 0.1 \times \frac{100}{1000} = 0.1 \times 0.1 = 0.01 \, \text{mol}.$$

For the second solution:

$$n_2=0.15\times \frac{100}{1000}=0.15\times 0.1=0.015\text{mol}\mathrm{.}$$

Step 2: Total moles of NaOH in the mixture

$$\text{Total moles} = n_1 + n_2 = 0.01 + 0.015 = 0.025 \, \text{mol}.$$

Step 3: Total volume of the mixture

$$\text{Total volume}=100{\text{cm}}^3+100{\text{cm}}^3=200{\text{cm}}^3=0.2{\text{dm}}^3\mathrm{.}$$

Step 4: Concentration of the resulting solution

Using the formula:

$$C=\frac{n}{V}<br>$$

where:

• $C$ = Concentration of the resulting solution ($\text{mol/dm}^3$),
• $n$ = Total moles (mol),
• $V$ = Total volume ($\text{dm}^3$).

Substitute the values:

$$C = \frac{0.025}{0.2} = 0.125 \, \text{mol/dm}^3.$$

Final Answer:

The concentration of the resulting solution is 0.125 mol/dm³.

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Question 8:

10 g of $\text{Na}_2\text{CO}_3$ are dissolved in water, and the solution is made up to $300 \, \text{cm}^3$. Find the concentration in moles per $\text{dm}^3$ of the solution.

$$\text{Na} = 23, \, \text{C} = 12, \, \text{O} = 16$$

Solution:

Step 1: Calculate the molar mass ​

$$\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \, \text{g/mol}.$$

Step 2: Calculate the number of moles​

Using the formula:

$$n=\frac{m}{M}<br>$$

where:

• $n$ = Number of moles (mol),
• $m$ = Mass of solute (g),
• $M$ = Molar mass (g/mol).

Substitute the values:

$$n = \frac{10}{106} \approx 0.09434 \, \text{mol}.$$

Step 3: Calculate the concentration in moles per dm3

Using the formula:

$$C=\frac{n}{V}<br>$$

where:

• $C$ = Concentration ($\text{mol/dm}^3$),
• $n$ = Number of moles (mol),
• $V$ = Volume of solution ($\text{dm}^3$).

Convert $300 \, \text{cm}^3$ to $\text{dm}^3$:

$$V = \frac{300}{1000} = 0.3 \, \text{dm}^3.$$

Substitute the values:

$$C = \frac{0.09434}{0.3} \approx 0.314 \, \text{mol/dm}^3.$$

Final Answer:

The concentration of the solution is approximately 0.314 mol/dm³.

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Question 9:

What mass of glucose $\text{C}_6\text{H}_{12}\text{O}_6$ is contained in 40 cm³ of $0.2 \, \text{mol/dm}^3$ solution?
Given:

$\text{[C}=12$[, $\text{H}=1$[, $\text{O}=16$].

Solution:

Step 1: Calculate the molar mass of glucose​

$${\text{Molar mass of C}}_6{\text{H}}_{12}{\text{O}}_6=(6\times 12)+(12\times 1)+(6\times 16)=72+12+96=180\text{g/mol}\mathrm{.}$$

Step 2: Calculate the number of moles of glucose

Using the formula:

$$n=C\times V,$$

where:

• $n$ = Number of moles (mol),
• $C$ = Concentration ($\text{mol/dm}^3$),
• $V$ = Volume ($\text{dm}^3$).

First, convert the volume from cm³ to dm³:

$$V=\frac{40}{1000}=0.04{\text{dm}}^3\mathrm{.}$$

Now, calculate the number of moles:

$$n = 0.2 \times 0.04 = 0.008 \, \text{mol}.$$

Step 3: Calculate the mass of glucose

Using the formula:

$$m = n \times M,$$

where:

• $m$ = Mass (g),
• $n$ = Number of moles (mol),
• $M$ = Molar mass (g/mol).

Substitute the values:

$$m = 0.008 \times 180 = 1.44 \, \text{g}.$$

Final Answer:

The mass of glucose in 40 cm³ of $0.2 \, \text{mol/dm}^3$ solution is 1.44 g.

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Applications of Molarity

1. Stoichiometry in Reactions: Molarity helps determine the exact amount of reactants and products in a chemical reaction.
2. Standard Solutions Preparation: Used in preparing solutions for titrations and other laboratory analyses.
3. Industrial Applications: Molarity is critical in industries for controlling concentrations in manufacturing processes like pharmaceuticals, food, and cleaning agents.
4. Medical Fields: Ensures accurate drug dosages in solutions such as saline or glucose drips.

Factors Affecting Molarity

1. Temperature: Molarity can change with temperature as the solution volume expands or contracts.
2. Purity of Solute: Impurities in the solute affect the accuracy of molarity.

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Mass Concentration

Introduction to Mass Concentration
Mass concentration is a way to express the concentration of a solute in a solution by relating the mass of the solute to the volume of the solution. It is commonly used in chemistry, medicine, and environmental science to describe the strength or density of a solution in terms of the amount of substance dissolved.

Definition of Mass Concentration

Mass concentration ($\rho$) is defined as the mass of solute ($m$) per unit volume of the solution ($V$). It is mathematically expressed as:

$$\rho =\frac{m}{V},$$

where:

• $\rho$ = Mass concentration (g/L or kg/m³),
• $m$ = Mass of the solute (g or kg),
• $V$ = Volume of the solution (L or m³).

Units of Mass Concentration

The most common units for mass concentration are:

1. Grams per liter ($g/L$)
2. Kilograms per cubic meter ($kg/m^3$)

How to Calculate Mass Concentration

To calculate the mass concentration of a solution:

1. Measure the mass of the solute: The quantity of the substance dissolved in the solvent.
2. Measure the volume of the solution: Ensure it is expressed in liters or cubic meters.
3. Apply the formula: $$\rho = \frac{m}{V}.$$

Examples of Mass Concentration Calculation

1. Example 1: Calculate the mass concentration of a solution containing 10 g of NaCl dissolved in 2 L of water.

   • Mass of solute ($m$) = 10 g,
   • Volume of solution ($V$) = 2 L.

   Substitute into the formula:

   $$\rho = \frac{m}{V} = \frac{10}{2} = 5 \, g/L.$$
   

2. Example 2: What is the mass concentration of a solution prepared by dissolving 5 kg of sugar in 0.01 $m^3$ of water?

   • Mass of solute ($m$) = 5 kg,
   • Volume of solution ($V$) = 0.01 $m^3$.

   Substitute into the formula:

   $$\rho = \frac{m}{V} = \frac{5}{0.01} = 500 \, kg/m^3.$$
   

Sample Questions and Answers on Mass Concentration

Question 1:

Calculate the concentration in g/dm³ of $0.75{\text{mol/dm}}^3$$\text{Na}_2\text{CO}_3$ solution.
Given: [$\text{Na}=23$, $\text{C}=12$, $\text{O}=16]$.

Solution:

Step 1: Calculate the molar mass of Na2​CO3​

$${\text{Molar mass of Na}}_2{\text{CO}}_3=(2\times 23)+12+(3\times 16)=46+12+48=106\text{g/mol}\mathrm{.}$$

Step 2: Calculate the concentration in g/dm³

Using the formula:

$${\text{Concentration (g/dm}}^3\text{)}={\text{Concentration (mol/dm}}^3\text{)}\times \text{Molar mass (g/mol)},$$

where:

• Concentration in mol/dm³ = 0.75 mol/dm³,
• Molar mass of $\text{Na}_2\text{CO}_3$ = 106 g/mol.

Substitute the values:

$$\text{Concentration (g/dm}^3\text{)} = 0.75 \times 106 = 79.5 \, \text{g/dm}^3.$$

Final Answer:

The concentration of the $0.75 \, \text{mol/dm}^3$ $\text{Na}_2\text{CO}_3$ solution is 79.5 g/dm³.

Question 2:

Calculate the value of $x$ in $\text{Ca(OH)}_2 \cdot x \text{H}_2\text{O}$ when the concentration in $\text{mol/dm}^3$ is 0.1 $\text{mol/dm}^3$ and the mass of the compound is 3.65 g in 250 cm³ of solution.
Given: [$\text{Ca }=40$, $\text{H }=1$, $\text{O }=16]$.

Solution:

Step 1: Calculate the moles of Ca(OH)2​⋅xH2​O in 250 cm³ of solution

First, convert the volume from cm³ to dm³:

$$V = \frac{250}{1000} = 0.25 \, \text{dm}^3.$$

Using the formula to calculate the number of moles:

$$n=C\times V,$$

where:

• $C = 0.1 \, \text{mol/dm}^3$,
• $V = 0.25 \, \text{dm}^3$.

Substitute the values:

$$n = 0.1 \times 0.25 = 0.025 \, \text{mol}.$$

Thus, the moles of $\text{Ca(OH)}_2 \cdot x \text{H}_2\text{O}$ in 250 cm³ of solution is 0.025 mol.

Step 2: Calculate the molar mass of Ca(OH)2​⋅xH2​O

To find the molar mass of $\text{Ca(OH)}_2 \cdot x \text{H}_2\text{O}$, we use the given mass of the compound (3.65 g) and the moles calculated in Step 1. The formula for molar mass is:

$$\text{Molar mass} = \frac{\text{Mass}}{\text{Moles}}.$$

Substitute the known values:

$$\text{Molar mass} = \frac{3.65}{0.025} = 146 \, \text{g/mol}.$$

Step 3: Calculate the molar mass of Ca(OH)2​ and determine the value of x

The molar mass of $\text{Ca(OH)}_2$ (without the water molecules) is calculated as follows:

$${\text{Molar mass of Ca(OH)}}_2=40+(2\times 16)+(2\times 1)=40+32+2=74\text{g/mol}\mathrm{.}$$

Now, subtract the molar mass of $\text{Ca(OH)}_2$ from the total molar mass of $\text{Ca(OH)}_2 \cdot x \text{H}_2\text{O}$ to find the contribution from the water molecules:

$$\text{Mass due to water} = 146 - 74 = 72 \, \text{g/mol}.$$

The molar mass of one water molecule ${\text{H}}_2\text{O is:}$

$$\text{Molar mass of H}_2\text{O} = (2 \times 1) + 16 = 18 \, \text{g/mol}.$$

Now, determine the number of water molecules $x$ by dividing the mass due to water by the molar mass of water:

$$x = \frac{72}{18} = 4.$$

Final Answer:

The value of $x$ is 4, so the compound is $\text{Ca(OH)}_2 \cdot 4 \text{H}_2\text{O}$.

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Question 3:

You are given a solution of $\text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O}$ in which 107.25 g of the salt is dissolved in 500 cm³ of solution. The concentration of the solution is 0.75 mol/dm³.
Calculate:
(a) The concentration in g/dm³ of the solution.
(b) The molar mass of the compound.
(c) The value of $x$.
Given: [$\text{Na }=23$, $\text{O }=16$, $\text{C }=12$, $\text{H }=1$].

Solution:

(a) The concentration in g/dm³ of the solution

The concentration in g/dm³ is given by:

$$\text{Concentration in g/dm}^3 = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (dm}^3\text{)}}.$$

First, convert 500 cm³ to dm³:

$$\text{Volume} = \frac{500}{1000} = 0.5 \, \text{dm}^3.$$

Now, calculate the concentration in g/dm³:

$$\text{Concentration in g/dm}^3 = \frac{107.25 \, \text{g}}{0.5 \, \text{dm}^3} = 214.5 \, \text{g/dm}^3.$$

Thus, the concentration is 214.5 g/dm³.

(b) The molar mass of the compound

To calculate the molar mass of the compound, we first use the given concentration (0.75 mol/dm³) and the volume of the solution (500 cm³ or 0.5 dm³) to find the moles of the compound:

$$\text{Moles of solute} = \text{Concentration} \times \text{Volume}.$$
$$\text{Moles of solute} = 0.75 \, \text{mol/dm}^3 \times 0.5 \, \text{dm}^3 = 0.375 \, \text{mol}.$$

Next, calculate the molar mass using the formula:

$$\text{Molar mass} = \frac{\text{Mass}}{\text{Moles}}.$$
$$\text{Molar mass} = \frac{107.25 \, \text{g}}{0.375 \, \text{mol}} = 285.0 \, \text{g/mol}.$$

Thus, the molar mass of the compound is 285.0 g/mol.

(c) The value of x

The formula for the compound is $\text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O}$. We know the molar mass of $\text{Na}_2\text{CO}_3$, and we can use it to determine $x$.

Molar mass of ${\text{Na}}_2{\text{CO}}_3$:

The molar mass of $\text{Na}_2\text{CO}_3$ is calculated as:

$$\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \, \text{g/mol}.$$
Molar mass of $x{\text{H}}_2\text{O}$:

Let the molar mass of water be $18 \, \text{g/mol}$. The contribution from $x$ water molecules is:

$$\text{Molar mass of } x \text{H}_2\text{O} = 18x \, \text{g/mol}.$$
Total molar mass:

The total molar mass of $\text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O}$ is:

$$\text{Molar mass of Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} = 106 + 18x \, \text{g/mol}.$$

We already calculated the molar mass to be 285.0 g/mol, so we set up the equation:

$$106 + 18x = 285.0.$$

Solving for $x$:

$$18x = 285.0 - 106 = 179.0, x = \frac{179.0}{18} = 9.94 \approx 10.$$

Thus, the value of $x$ is 10.

Final Answer:

(a) The concentration in g/dm³ is 214.5 g/dm³.
(b) The molar mass of the compound is 285.0 g/mol.
(c) The value of $x$ is 10, so the formula is $\text{Na}_2\text{CO}_3 \cdot 10 \text{H}_2\text{O}$.

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Question 4:

9.0g of the hydrated salt $\text{CuSO}_4 \cdot x \text{H}_2\text{O}$was dissolved in 500 cm³ of water, and the concentration in mol/dm³ was experimentally found to be 0.072 mol/dm³.
Calculate:
(i) The molar mass of $\text{CuSO}_4 \cdot x \text{H}_2\text{O}$.
(ii) The value of $x$, the number of water molecules of crystallization.
Given: [Cu = 64, $S\; =32$, $\text{O }=16$, $\text{H }=1]$.

Solution:

(i) The molar mass of CuSO4​⋅xH2​O

To calculate the molar mass, we first use the formula for molar mass:

$$\text{Molar mass} = \frac{\text{Mass of solute}}{\text{Moles of solute}}.$$

We are given the mass of the hydrated salt $\text{CuSO}_4 \cdot x \text{H}_2\text{O}$ as 9.0g and the concentration of the solution as 0.072 mol/dm³. The volume of the solution is 500 cm³, which is equal to 0.5 dm³.

Moles of ${\text{CuSO}}_4\cdot x{\text{H}}_2\text{O}$:
$$\text{Moles of solute} = \text{Concentration} \times \text{Volume}.$$
$$\text{Moles of solute} = 0.072 \, \text{mol/dm}^3 \times 0.5 \, \text{dm}^3 = 0.036 \, \text{mol}.$$
Molar mass of ${\text{CuSO}}_4\cdot x{\text{H}}_2\text{O}$: Now, using the formula for molar mass:
$$\text{Molar mass} = \frac{9.0 \, \text{g}}{0.036 \, \text{mol}} = 250 \, \text{g/mol}.$$

Thus, the molar mass of $\text{CuSO}_4 \cdot x \text{H}_2\text{O}$ is 250 g/mol.

(ii) The value of x, the number of water molecules of crystallization

To calculate the value of $x$, we need to break down the molar mass of the hydrated salt into two parts:

The molar mass of ${\text{CugO}}_4$.
The molar mass contribution from $x$ water molecules.

Molar mass of ${\text{CuSO}}_4$: The molar mass of ${\text{CuSO}}_4$ is:
$$\text{Molar mass of CuSO}_4 = 64 + 32 + (4 \times 16) = 64 + 32 + 64 = 160 \, \text{g/mol}.$$
Molar mass contribution from water $x{\text{H}}_2\text{O}$: The molar mass of one water molecule is 18 g/mol, so the molar mass contribution from $x$ water molecules is:
$$\text{Molar mass of water} = 18x \, \text{g/mol}.$$
Total molar mass of ${\text{CuSO}}_4\cdot x{\text{H}}_2\text{O}$: The total molar mass of ${\text{CuSO}}_4\cdot x{\text{H}}_2\text{O}$ is:
$$\text{Total molar mass} = 160 + 18x.$$

We already know that the total molar mass is 250 g/mol, so we can set up the equation:

$$160 + 18x = 250.$$

Solving for $x$:

$$18x = 250 - 160 = 90,$$
$$x=\frac{90}{18}=5.$$

Thus, the value of $x$ is 5.

Final Answer:

(i) The molar mass of $\text{CuSO}_4 \cdot x \text{H}_2\text{O}$ is 250 g/mol.
(ii) The value of $x$ is 5, so the formula of the hydrated salt is $\text{CuSO}_4 \cdot 5 \text{H}_2\text{O}$.

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Applications of Mass Concentration

1. Environmental Science: Used to measure pollutant concentrations, such as $\text{mg/L}$ for contaminants in water or air.

2. Pharmaceuticals: Mass concentration is vital in formulating drugs and intravenous solutions to ensure proper dosages.

3. Food Industry: Used to determine the concentration of ingredients, such as sugar or salt in beverages or processed foods.

4. Industrial Processes: Helps in mixing chemicals for production, ensuring consistency and efficiency.

Factors Affecting Mass Concentration

1. Temperature: Affects the solution's volume due to thermal expansion, thereby influencing the concentration.
2. Precision of Measurements: Errors in measuring the solute's mass or the solution's volume can affect the calculated concentration.

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Concentration in Terms of Parts per Million (PPM)

Parts per million (PPM) is a unit of concentration commonly used to express the amount of a substance (typically a contaminant or solute) in a solution, typically in water or air. It represents the ratio of one part of a substance to one million parts of the solution, which can be expressed as:

$$\text{PPM} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 10^6$$

Explanation:

• PPM is used when the solute is present in very low concentrations, such as pollutants in water or trace elements in air.
• 1 PPM means 1 part of the solute for every 1 million parts of the solution. In practice, this is often equivalent to 1 milligram of solute per liter of solution (mg/L), though the exact conversion can depend on the density of the solution.

How to Calculate PPM:

• For liquid solutions: PPM can be calculated as:

$$\text{PPM}=\frac{\text{Mass of solute (mg)}}{\text{Volume of solution (L)}}\times 10^3$$

This equation assumes that the density of the solution is close to that of water (1 g/cm³).

• For gases: For gases in air, PPM can also be calculated using the ratio of the number of moles of the substance to the number of moles of air.

Example:

Problem 1:
If you have a solution where 0.02g of sodium chloride (NaCl) is dissolved in 5000g of water, calculate the concentration in PPM.

Solution:

1. Mass of solute (NaCl): 0.02g
2. Mass of solution (water + NaCl): 5000g + 0.02g = 5000.02g

Now, use the formula:

$$\text{PPM} = \frac{0.02 \, \text{g}}{5000.02 \, \text{g}} \times 10^6$$
$$\text{PPM} = \frac{0.02}{5000.02} \times 10^6 = 4.00 \, \text{PPM}.$$

Thus, the concentration of NaCl in the solution is 4.00 PPM.

Problem 2:

Convert 0.05% of NaCl solution to ppm

Solution:

To convert the concentration of a NaCl solution from percentage (% weight/volume) to parts per million (ppm), we can use the following relationship:

$$\text{ppm} = \text{Percentage} \times 10,000$$

Given:

• Concentration of NaCl solution = 0.05%

$$\text{ppm} = 0.05\% \times 10,000$$
$$\text{ppm}=500\text{ppm}$$

Answer:

The concentration of the NaCl solution is 500 ppm.

Problem 3:

A pesticide solution contains 0.4g of the active substance in 8 dm³. What is the concentration in ppm?

Solution:

o calculate the concentration in parts per million (ppm), we can use the following formula:

$$\text{ppm}=\frac{\text{Mass of solute (g)}}{{\text{Volume of solution (dm}}^3\text{)}}\times 10^6$$

Given:

• Mass of active substance = 0.4 g
• Volume of solution = 8 dm³

$$\text{ppm} = \frac{0.4 \, \text{g}}{8 \, \text{dm}^3} \times 10^6$$
$$\text{ppm} = \frac{0.4}{8} \times 10^6$$
$$\text{ppm} = 0.05 \times 10^6$$
$$\text{ppm}=50,000\text{ppm}$$

Answer:

The concentration of the pesticide solution is 50,000 ppm.

Problem 4:

The concentration of an aqueous solution is 5mg.dm⁻³ to parts. What is the concentration in part per million (ppm). 

Solution:

To convert the concentration from mg.dm⁻³ to parts per million (ppm), we can use the fact that:

$$1{\text{mg.dm}}^{-3}=1\text{ppm}$$

Given:

• Concentration = 5 mg.dm⁻³

Since 1 mg.dm⁻³ is equal to 1 ppm, we can directly state:

$$\text{Concentration in ppm}=5\text{ppm}$$

Answer:

The concentration is 5 ppm.

Key Points to Remember:

• 1 PPM = 1 mg of solute per liter of solution for water-based solutions.
• For solid samples, PPM is calculated using mass ratios.
• PPM is commonly used in environmental science, water quality testing, and chemical analysis.

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Molality (m)

Molality is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is based on the volume of the solution, molality is based on the mass of the solvent, making it useful for studying colligative properties (e.g., boiling point elevation, freezing point depression), as these properties are independent of the volume of the solution.

Molality Formula:

$$\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$$

Where:

• Moles of solute is the amount of the solute in moles.
• Mass of solvent is the mass of the solvent in kilograms (not the total solution).

Units of Molality:

• The unit of molality is mol/kg (moles per kilogram).

Example Problem:

Problem 1:
How many moles of NaCl are present in 3 kg of water if the molality of the NaCl solution is 2 mol/kg?

Solution: Given:

• Molality $m = 2 \, \text{mol/kg}$
• Mass of solvent (water) = 3 kg

Using the formula:

$$m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$$

Rearranging for moles of solute:

$$\text{Moles of solute} = m \times \text{Mass of solvent}$$
$$\text{Moles of solute} = 2 \, \text{mol/kg} \times 3 \, \text{kg} = 6 \, \text{moles of NaCl}$$

Thus, there are 6 moles of NaCl in the solution.

Problem 2:

Calculate the molality of 8g of NaOH dissolved in 400g of water. [Na = 23, O = 16, H = 1]

Solution:

The formula for molality ($m$) is:

$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$

Step 1: Calculate the moles of NaOH.

To calculate the moles of NaOH, we use the formula:

$$\text{moles}=\frac{\text{mass}}{\text{molar mass}}$$

The molar mass of NaOH is:

$$\text{Molar mass of NaOH} = 23 \, (\text{Na}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 40 \, \text{g/mol}$$

Now, calculate the moles of NaOH:

$$\text{moles of NaOH} = \frac{8 \, \text{g}}{40 \, \text{g/mol}} = 0.2 \, \text{mol}$$

Step 2: Calculate the mass of the solvent (water) in kg.

Given that the mass of water is 400g, we convert it to kg:

$$400\text{g}=0.4\text{kg}$$

Step 3: Calculate the molality.

Now we can calculate the molality:

$$m=\frac{0.2\text{mol}}{0.4\text{kg}}=0.5\text{mol/kg}$$

Answer:

The molality of the NaOH solution is 0.5 mol/kg.

Applications of Molality:

1. Colligative Properties: Molality is used to calculate the changes in freezing point and boiling point of a solution. For example, the freezing point depression is proportional to the molality of the solution.
2. Temperature Independence: Since molality depends on mass (which doesn't change with temperature), it is more accurate in experiments that involve significant temperature changes compared to molarity, which depends on volume (which can change with temperature).

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Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a component to the total number of moles of all components in the mixture.

Mathematically, the mole fraction of a component $A$ (denoted $\chi_A$) in a mixture is given by:

$${\chi }_A=\frac{n_A}{n_{\text{total<span class="vlist"><span class="mord"><span class="mord"><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist"></span></span></span></span></span></span></span><span class="vlist-s"></span>}}}$$

Where:

• $\chi_A$ = Mole fraction of component A
• $n_A$ = Number of moles of component A
• $n_{\text{total}}$ = Total number of moles of all components in the mixture

Mole Fraction of Solvent (Water)

Similarly, for solvent $B$ (for example, water), the mole fraction $\chi_B$ is:

$${\chi }_B=\frac{n_B}{n_{\text{total<span class="vlist-s"></span>}}}$$

Where:

• $\chi_B$ = Mole fraction of component B (e.g., water)
• $n_B$ = Number of moles of component B (e.g., water)
• $n_{\text{total}}$ = Total number of moles of all components in the mixture

Steps to Calculate Mole Fraction:

1. Determine the moles of each component:

• Use the formula $n=\frac{m}{M}$, where:
  • $m$ = mass of the substance
  • $M$ = molar mass of the substance

2. Find the total number of moles in the mixture:

• Add the moles of all components.

3. Calculate the mole fraction of the component of interest:

• Divide the moles of the component by the total number of moles.

Example Calculation:

Question 1:

Given:

• 4g of NaCl (molar mass of NaCl = 58.5 g/mol)
• 50g of water (molar mass of H₂O = 18 g/mol)

Step 1: Calculate the moles of NaCl:

$$n_{\text{NaCl}} = \frac{4 \, \text{g}}{58.5 \, \text{g/mol}} = 0.068 \, \text{mol}$$

Step 2: Calculate the moles of water:

$$n_{\text{H}_2\text{O}} = \frac{50 \, \text{g}}{18 \, \text{g/mol}} = 2.78 \, \text{mol}$$

Step 3: Find the total moles:

$$n_{\text{total}} = n_{\text{NaCl}} + n_{\text{H}_2\text{O}} = 0.068 \, \text{mol} + 2.78 \, \text{mol} = 2.848 \, \text{mol}$$

Step 4: Calculate the mole fraction of NaCl:

$$\chi_{\text{NaCl}} = \frac{n_{\text{NaCl}}}{n_{\text{total}}} = \frac{0.068}{2.848} = 0.0239$$

Step 5: Calculate the mole fraction of water:

$$\chi_{\text{H}_2\text{O}} = \frac{n_{\text{H}_2\text{O}}}{n_{\text{total}}} = \frac{2.78}{2.848} = 0.9761$$

Answer:

• The mole fraction of NaCl is 0.0239.
• The mole fraction of water is 0.9761.

Question 2:

Calculate the mole fraction of oxygen in a mixture of 8g of oxygen, 11g of carbon (IV) oxide, and 3.5g of nitrogen gas at room temperature (25°C). [O = 16, C = 12, N = 14]

Solution:

To calculate the mole fraction of oxygen in the mixture, we need to follow these steps:

1. Calculate the number of moles of oxygen (O₂).
2. Calculate the number of moles of carbon dioxide (CO₂).
3. Calculate the number of moles of nitrogen (N₂).
4. Find the total number of moles in the mixture.
5. Calculate the mole fraction of oxygen.

Step 1: Calculate the moles of oxygen (O₂):

The molar mass of oxygen (O₂) is 32 g/mol.

$$n_{\text{O}_2} = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = 0.25 \, \text{mol}$$

Step 2: Calculate the moles of carbon dioxide (CO₂):

The molar mass of carbon dioxide (CO₂) is:

$$M_{\text{CO}_2} = 12 + 2(16) = 44 \, \text{g/mol}$$

Now, calculate the moles of CO₂:

$$n_{\text{CO}_2} = \frac{11 \, \text{g}}{44 \, \text{g/mol}} = 0.25 \, \text{mol}$$

Step 3: Calculate the moles of nitrogen (N₂):

The molar mass of nitrogen (N₂) is:

$$M_{{\text{N}}_2}=2(14)=28\text{g/mol}$$

Now, calculate the moles of N₂:

$$n_{\text{N}_2} = \frac{3.5 \, \text{g}}{28 \, \text{g/mol}} = 0.125 \, \text{mol}$$

Step 4: Find the total number of moles in the mixture:

$$n_{\text{total}} = n_{\text{O}_2} + n_{\text{CO}_2} + n_{\text{N}_2}$$ $$n_{\text{total}} = 0.25 \, \text{mol} + 0.25 \, \text{mol} + 0.125 \, \text{mol} = 0.625 \, \text{mol}$$

Step 5: Calculate the mole fraction of oxygen (O₂):

The mole fraction of oxygen is given by:

$$\chi_{\text{O}_2} = \frac{n_{\text{O}_2}}{n_{\text{total}}}$$ $${\chi }_{{\text{O}}_2}=\frac{0.25\text{mol}}{0.625\text{mol}}=0.4$$

Answer:

The mole fraction of oxygen (O₂) in the mixture is 0.4.

Question 3:

Calculate the mole fraction of NaOH in a mixture containing 8g of NaOH dissolved in 1.7g of NH₃ and 5.3g of Na₂CO₃. [Na = 23, O = 16, H = 1, C = 12, N = 14]

Solution:

To calculate the mole fraction of NaOH in the mixture, we need to follow these steps:

1. Calculate the moles of NaOH.
2. Calculate the moles of NH₃.
3. Calculate the moles of Na₂CO₃.
4. Find the total number of moles in the mixture.
5. Calculate the mole fraction of NaOH.

Step 1: Calculate the moles of NaOH:

The molar mass of NaOH is:

$$M_{\text{NaOH}} = 23 \, (\text{Na}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 40 \, \text{g/mol}$$

Now, calculate the moles of NaOH:

$$n_{\text{NaOH}} = \frac{8 \, \text{g}}{40 \, \text{g/mol}} = 0.2 \, \text{mol}$$

Step 2: Calculate the moles of NH₃:

The molar mass of NH₃ is:

$$M_{\text{NH}_3} = 14 \, (\text{N}) + 3 \, (\text{H}) = 17 \, \text{g/mol}$$

Now, calculate the moles of NH₃:

$$n_{\text{NH}_3} = \frac{1.7 \, \text{g}}{17 \, \text{g/mol}} = 0.1 \, \text{mol}$$

Step 3: Calculate the moles of Na₂CO₃:

The molar mass of Na₂CO₃ is:

$$M_{\text{Na}_2\text{CO}_3} = 2(23 \, \text{Na}) + 12 \, \text{C} + 3(16 \, \text{O}) = 46 + 12 + 48 = 106 \, \text{g/mol}$$

Now, calculate the moles of Na₂CO₃:

$$n_{\text{Na}_2\text{CO}_3} = \frac{5.3 \, \text{g}}{106 \, \text{g/mol}} = 0.05 \, \text{mol}$$

Step 4: Find the total number of moles in the mixture:

$$n_{\text{total}} = n_{\text{NaOH}} + n_{\text{NH}_3} + n_{\text{Na}_2\text{CO}_3}$$ $$n_{\text{total}} = 0.2 \, \text{mol} + 0.1 \, \text{mol} + 0.05 \, \text{mol} = 0.35 \, \text{mol}$$

Step 5: Calculate the mole fraction of NaOH:

The mole fraction of NaOH is given by:

$$\chi_{\text{NaOH}} = \frac{n_{\text{NaOH}}}{n_{\text{total}}}$$ $$\chi_{\text{NaOH}} = \frac{0.2 \, \text{mol}}{0.35 \, \text{mol}} = 0.571$$

Answer:

The mole fraction of NaOH in the mixture is 0.571.

Question 4:

2g of AgNO₃ is dissolved in 100g of water. Calculate the:

1. Molality of the solution
2. Mole fraction of AgNO₃ in the solution
   [Mr AgNO₃ = 170]

Solution:

To calculate both the molality and the mole fraction of AgNO₃, we will follow the steps below:

Step 1: Calculate the moles of AgNO₃.

The molar mass of AgNO₃ is:

$$M_{{\text{AgNO}}_3}=107.87(\text{Ag})+14.01(\text{N})+3(16\text{O})=170\text{g/mol}$$

Now, calculate the moles of AgNO₃:

$$n_{{\text{AgNO}}_3}=\frac{2\text{g}}{170\text{g/mol}}=0.01176\text{mol}$$

Step 2: Calculate the molality.

Molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{n_{\text{AgNO}_3}}{m_{\text{water}}}$$

Where the mass of the water is 100g, which is equal to 0.1 kg.

$$m=\frac{0.01176\text{mol}}{0.1\text{kg}}=0.1176\text{mol/kg}$$

Step 3: Calculate the mole fraction of AgNO₃.

The mole fraction of AgNO₃ is given by the formula:

$$\chi_{\text{AgNO}_3} = \frac{n_{\text{AgNO}_3}}{n_{\text{AgNO}_3} + n_{\text{water}}}$$

First, we need to calculate the moles of water. The molar mass of water (H₂O) is:

$$M_{\text{H}_2\text{O}} = 2(1) + 16 = 18 \, \text{g/mol}$$

Now, calculate the moles of water:

$$n_{\text{water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.56 \, \text{mol}$$

Now we can calculate the mole fraction:

$$\chi_{\text{AgNO}_3} = \frac{0.01176 \, \text{mol}}{0.01176 \, \text{mol} + 5.56 \, \text{mol}} = \frac{0.01176}{5.57176} = 0.00211$$

Answer:

1. Molality of the solution = 0.1176 mol/kg
2. Mole fraction of AgNO₃ = 0.00211

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Practical Applications of the Mole Concept

Understanding moles is not just crucial for academic studies. The mole concept plays a vital role in real-world applications:

• Pharmaceuticals: Calculating drug dosages based on molecular weight and molarity.
• Industry: Determining the yields of chemical reactions in manufacturing processes.
• Environmental Science: Measuring the concentration of pollutants in the air or water.

Conclusion

The amount of substance and the mole are core concepts in chemistry that allow scientists to quantify the number of particles in a substance, simplify complex calculations, and make accurate predictions in chemical reactions. By understanding how to convert between moles, mass, volume, and number of particles, and by mastering concepts like molar mass, stoichiometry, and molarity, students and professionals can apply chemistry principles in practical, real-world situations.

Key Takeaways:

• The mole helps simplify the complexity of chemistry by bridging the atomic and macroscopic scales.
• Avogadro's number ($6.022 \times 10^{23}$) is a crucial constant in chemistry.
• Molar mass, molar volume, and stoichiometry are essential tools in chemical calculations.
• Real-world applications of the mole concept are seen in pharmaceuticals, environmental science, and industrial chemistry.

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