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

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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 (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.

 

Check Out: Sample Questions and Answers on Molar Volume

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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)

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

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.

Check Out: Sample Questions and Answers on 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:

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

Applications of Mass Concentration

1. Environmental Science: Used to measure pollutant concentrations, such as 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.

Learn More on: Sample Questions and Answers on Mass 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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