Question

The volume of $3.7g$ of a gas at ${25}^{o}C$ is the same as the volume of $0.184g$ of $H_2$ at ${17}^{o}C$ and same pressure. What is the molecular weight of the gas?

• A. $31.33g/mol$
• B. $41.33g/mol$
• C. $21.45g/mol$
• D. $11.45g/mol$

Answer 1

Answer: (B)

$41.33g/mol$

From ideal gas equation-

$PV=nRT$

whereas,

$P=$ pressure of the gas

$V=$ volume it occupies

$n=$ number of moles of gas present in the sample

$R=$ universal gas constant $=0.0821atmL{mol}^{-1}{K}^{-1}$

$T=$ absolute temperature of the gas

At constant pressure and volume,

$n\propto \frac{1}{T}$

$\therefore \frac{n_1}{n_2}=\frac{T_2}{T_1}$

$\Rightarrow n_2=\frac{n_1\times T_1}{T_2}$

Let $n_1$ and $n_2$ be the no. of moles of $H_2$ and the other gas respectively.

Let $T_1$ and $T_2$ be the temperature for $H_2$ and the other gas respectively.

Let the molecular weight of other gas be 'M' g.

Given weight of $H_2=0.194g$

Molecular wt. of $H_2=2g$

Given wt. of other gas $=3.7g$

Molecular weight of other gas $=M=?$

As we know that,

$\text{no. of moles}=\frac{\text{given wt.}}{\text{mol. wt.}}$

Therefore,

$n_1=$ moles of $H_2=\frac{0.184}{2}=0.092\text{mole}$

$n_2=$ moles of other gas $=\frac{3.7}{M}$

$T_1=17℃=(17+273)K=290K$

$T_2=25℃=(25+273)K=298K$

$\therefore \frac{3.7}{M}=\frac{0.092\times 290}{298}$

$\Rightarrow M=\frac{3.7\times 298}{0.092\times 290}=41.3268g\approx 41.33g$

Hence the molecular weight of other gas is 41.33 g.