Introduction:

Molar volume is a crucial concept in chemistry, particularly when studying gases. It refers to the volume occupied by one mole of a gas at standard temperature and pressure (STP). Understanding how to calculate and apply molar volume can help students excel in various problem-solving scenarios. In this post, we’ll explore sample questions and answers that illustrate the practical use of molar volume in solving problems related to gas laws.

Sample questions and answers that illustrate the practical use of molar volume

How to Calculate Molar Volume

Molar volume refers to the volume occupied by one mole of a substance, typically a gas, at specific conditions of temperature and pressure. For gases, molar volume is most commonly calculated under standard temperature and pressure (STP) or room temperature and pressure (RTP). Here’s a step-by-step guide to calculating molar volume:

1. Understand the Conditions

At STP (Standard Temperature and Pressure):
- Temperature = 273.15 K (0°C)
- Pressure = 1 atm (101.3 kPa)
- Molar volume = 22.4 L/mol (for ideal gases)
At RTP (Room Temperature and Pressure):
- Temperature = 298 K (25°C)
- Pressure = 1 atm
- Molar volume ≈ 24.0 L/mol

Note: These values are approximations and assume ideal gas behavior.

2. Use the Ideal Gas Law

The molar volume can be derived using the Ideal Gas Law:
$PV = nRT$

Where:

$P$ = pressure (in atm or Pa)
$V$ = volume (in liters or cubic meters)
$n$ = number of moles
$R$ = ideal gas constant $(0.0821 \, L·atm·mol^{-1}·K^{-1} \, \text{or} \, 8.314 \, J·mol^{-1}·K^{-1})$
$T$ = temperature (in Kelvin)

For one mole of a gas ( $n = 1$ ):
$V = \frac{RT}{P}$

3. Plug in the Values

At STP:
$V = \frac{(0.0821 \, L·atm·mol^{-1}·K^{-1})(273.15 \, K)}{1 \, atm} = 22.4 \, L/mol$
At RTP:
$V = \frac{(0.0821 \, L·atm·mol^{-1}·K^{-1})(298 \, K)}{1 \, atm} ≈ 24.0 \, L/mol$

4. Adjust for Non-Ideal Conditions

If the gas does not behave ideally, corrections may be required using the Van der Waals equation or other modifications. This is especially true at very high pressures or low temperatures.

5. Example Problem

Question: Calculate the volume occupied by 2 moles of oxygen gas at STP.

Solution:

Given: $n = 2 \, \text{mol}, \, P = 1 \, atm, \, T = 273.15 \, K$
Use $V = \frac{nRT}{P}$ :
$V = \frac{(2 \, \text{mol})(0.0821 \, L·atm·mol^{-1}·K^{-1})(273.15 \, K)}{1 \, atm}$
$V = 44.8 \, L$

Thus, 2 moles of oxygen gas occupy 44.8 liters at STP.

Sample Questions and Answers on Molar Volume

Question 1:

Calculation of Amount of CO₂ in 28 dm³ at STP

Given:

Molar Volume ($ V_m $) = 22.4 dm³/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 dm³ of gas at STP is 1.25 moles.

Question 2:

Calculation of Relative Molecular Mass of Trichloromethane (CHCl₃)
Given:
Mass of trichloromethane = 0.12 g
Volume of vapor formed = 22.4 cm³
Molar volume at STP = 22.4 dm³ = 22,400 cm³
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}} $
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 the formula:
$ 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.

Question 3:

Calculation of Moles, Volume, and Molecules of Diethyl Ether (C₂H₅)₂O

Given:

Mass of diethyl ether = 3.7 g
Molar mass of diethyl ether (C₂H₅)₂O = 74 g/mol
Avogadro's number = $6.02 \times 10^{23}$ mol$^{-1}$
Molar volume at STP = 22.4 dm³/mol

Solution:

(a) Calculate the number of moles of diethyl ether (C₂H₅)₂O

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

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

Substitute the given values:

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

Simplify:

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

(b) Calculate the volume of the vapor at STP

The volume ($ V $) of the vapor at STP 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 diethyl ether (C₂H₅)₂O

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 (C₂H₅)₂O is $ 0.05 \, \text{mol} $
(b) The volume of the vapor at STP is $ 1.12 \, \text{dm}^3 $
(c) The number of molecules of (C₂H₅)₂O is $ 3.01 \times 10^{22} $ molecules

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 ( $V_m$ ) = $22.4 {dm}^{3} / mol$
Mass = $11 g$

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.

Question 5:

Calculation of Molar Mass and Atomic Mass of XX in X₂O₅

Given:

Volume ($ V $) = 2.8 dm³
Molar volume ($ V_m $) = 22.4 dm³/mol
Mass = 17.75 g
Oxygen atomic mass ($ O $) = 16 g/mol

Solution:

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

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

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

Substitute the given 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 $ 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}) $

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 \, \text{g/mol} $
The atomic mass of $ X $ is $ 31 \, \text{g/mol} $

Question 6:

Calculation of Molar Mass and Atomic Mass of X in X₂O₃

Given:

Volume ($ V $) = 5.6 dm³
Molar volume ($ V_m $) = 22.4 dm³/mol
Mass = 16 g
Oxygen atomic mass ($ O $) = 16 g/mol

Solution:

Step 1: Calculate the number of moles of $ X_2O_3 $

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

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

Substitute the given values:

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

Simplify:

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

Step 2: Calculate the molar mass of $ X_2O_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 $ X_2O_3 $ can also be written 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:

The molar mass of $ X_2O_3 $ is $ 64 \, \text{g/mol} $
The atomic mass of $ X $ is $ 8 \, \text{g/mol} $

Question 7:

Calculation of Molar Mass and Value of x in CₓH₈

Given:

Volume ($ V $) = 2.8 dm³
Molar volume ($ V_m $) = 22.4 dm³/mol
Mass = 7 g
Carbon atomic mass ($ C $) = 12 g/mol
Hydrogen atomic mass ($ H $) = 1 g/mol

Solution:

Step 1: Calculate the number of moles of CₓH₈

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

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

Substitute the given values:

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

Simplify:

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

Step 2: Calculate the relative molar mass (M) of CₓH₈

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 CₓH₈ 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:

The relative molar mass of CₓH₈ is $ 56 \, \text{g/mol} $
The value of $ x $ is $ 4 $, so the molecular formula is $ C_4H_8 $

Question 8:

Calculation of Molar Mass and Value of $ x $ in $ (C_2H_5)_xO $

Given:

Volume ($ V $) = 1.12 dm³
Molar volume ($ V_m $) = 22.4 dm³/mol
Mass = 3.7 g
Atomic masses: $ C = 12 $, $ H = 1 $, $ 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 given 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 calculated as:

$ (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:

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

Conclusion:

Mastering the concept of molar volume is essential for any chemistry student, as it forms the foundation for understanding gas behavior under different conditions. By practicing sample questions and answers, you can develop the skills needed to solve real-world chemistry problems efficiently. With this knowledge, you’ll be better equipped to tackle complex calculations and apply the ideal gas law confidently in your studies.

SHS Chemistry Topics

Sample Questions and Answers on Molar Volume

Introduction:

How to Calculate Molar Volume

1. Understand the Conditions

2. Use the Ideal Gas Law

3. Plug in the Values

4. Adjust for Non-Ideal Conditions

5. Example Problem

Sample Questions and Answers on Molar Volume

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Calculation of Molar Mass and Value of x in CₓH₈

Question 8:

Conclusion:

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Sample Questions and Answers on Molar Volume

Introduction:

How to Calculate Molar Volume

1. Understand the Conditions

2. Use the Ideal Gas Law

3. Plug in the Values

4. Adjust for Non-Ideal Conditions

5. Example Problem

Sample Questions and Answers on Molar Volume

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Calculation of Molar Mass and Value of x in CₓH₈

Question 8:

Conclusion:

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