Question

A vessel of 25 L contains 20 g of an ideal gas X at 300K. The pressure exerted by the gas is 1 atm. 20 g of ideal gas Y is added to the vessel keeping the temperature constant. The total pressure became 3 atm. Upon further addition of 20 g ideal gas Z, the pressure became 7 atm. Find the molar mass of gas X.
[Take R = 1/12 $L-atmmo{l}^{-1}{K}^{-1}$]

• A. 20 g
• B. 10 g
• C. 30 g
• D. 5 g

Answer 1

The correct option is A 20 g

The ideal gas equation :

$PV=nRT$

$=\frac{m}{M}RT$

Given values for ideal gas $X$,

$V=25L$

$m=20g$

$T=300K$

$P=1atm$

Substitute values in the above equation :

$1\times 25=\frac{20}{M}\times \frac{1}{12}\times 300$.
$M=\frac{20\times 300}{25\times 12}$

Thus, the molar mass is $M=20g/mol$.
Hence, option $\left(A\right)$ is correct.