Chapter 13.2, Problem 23PP

Interpretation Introduction

Interpretation:

The volume of the gas is to be calculated at STP for a given mass of the gas.

Concept introduction:

Ideal gas equation is used to calculate volume of the gas at given temperature and pressure.

According to this law,

$\mathrm{PV}=nRT$

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Answer to Problem 23PP

The volume of the gas is $0.111L$

Explanation of Solution

Given information:

The mass of krypton $\mathrm{m}=0.416\mathrm{g}$

Ideal gas equation helps to calculate volume of a gas at given temperature and pressure. According to this law,

$\mathrm{PV}=nRT$

Or

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

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Now

Given mass of gas is $0.416g$

At STP

Pressure is $1atm$

Temperature is $273K$

Gas constant is $0.0821Latm{K}^{-1}mo{l}^{-1}$

Molar mass is defined as average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The given gas is $\mathrm{Kr}$.

The formula to calculate molar mass is,

$\mathrm{Molar}mass=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• ${\mathrm{N}}_1$ represents the number of atoms of first element in chemical formula.
• ${\mathrm{x}}_1$ represents the atomic mass of first element.
• ${\mathrm{N}}_2$ represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of second element.

Atomic mass of Krypton is $83.8gmo{l}^{-1}$

Hence,

$\begin{array}{c}M=1\times 83.8\\ =83.8gmo{l}^{-1}\end{array}$

Further number of moles $\mathrm{n}$ is calculated by using formula:

$\begin{array}{c}\mathrm{n}=\frac{m}{M}\\ =\frac{0.416}{83.8}\\ =0.0049mol\end{array}$

$\mathrm{m}$ denotes mass of gas

$\mathrm{M}$ denotes molar mass of gas

On substituting the values in the formula:

$\begin{array}{c}V=\frac{0.0049\times 0.0821\times 273}{1}\\ =0.111L\end{array}$

Therefore, volume of the gas is $0.111L$

Conclusion

The volume of the gas at STP is $0.111L$.