Chapter 13.2, Problem 21PP

Interpretation Introduction

Interpretation:

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

Concept introduction:

Ideal gas equation is used to calculate mass 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 21PP

The mass of the gas is $1.96g$

Explanation of Solution

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

$\mathrm{PV}=nRT$

Or

$\mathrm{n}=\frac{PV}{RT}$

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

Further, mass of gas $\mathrm{m}$ is calculated by multiplying number of moles $\mathrm{n}$ with molar mass of the gas $\mathrm{M}$

Now volume is $1L$

At STP

Pressure is $1atm$

Temperature is $273K$

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

On substituting the values in the formula:

$\begin{array}{c}\mathrm{n}=\frac{1\times 1}{\begin{array}{l}0.0821\times 273\\ \end{array}}\\ =0.044g\end{array}$

So $\mathrm{m}$ is given by

$\mathrm{m}=M\times n$

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 ${\mathrm{CO}}_2$.

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 carbon is $12gmo{l}^{-1}$

Atomic mass of oxygen is $16gmo{l}^{-1}$

Hence,

$\begin{array}{c}M=1\times 12+2\times 16\\ =12+32\\ =44gmo{l}^{-1}\end{array}$

$\begin{array}{c}\mathrm{m}=44\times 0.044\\ =1.96g\end{array}$

Therefore, mass of the gas is $1.96g$

Conclusion

The mass of the gas at STPis $1.96g$.