5.0g of a mixture of He and another gas occupies a volume of 1.5 L at   300 K and 750 mm Hg. The gas freezes at 270 K. At 15 K the pressure of the gas mixture is 8 mm Hg at the same volume. What is the molar mass of the gas ?

According to ideal gas equation,

$\text{PV}=\text{nRT}$

From the given data of a mixture,

$\frac{750}{760}\times 1.5=n\times 0.0821\times 300$

$\text{Note:}760\text{mm of}\text{Hg}=1\text{atm}$

$n_{mix}=\frac{750\times 1.5}{760\times 24.63}$

$n_{\text{mixure}}=0.06\text{moles}$

Volume of He gas at 15 K = Volume of mixture of gaess at 300 K

At 15K, He is present in the form of gas.

Total pressure of mixture is equal to pressure of He gas.

${\text{P}}_{\text{He}}=8\text{mm}\text{of}\text{Hg}\text{=}\frac{8}{760}\text{atm}$

Apply $\text{PV}=\text{nRT}$   for He gas at 15 K.

$\frac{8}{760}\times 1.5=n_{He}\times 0.0821\times 15$

$n_{He}=\frac{8}{760\times 0.0821}\times \frac{1.5}{15}=0.0128\text{mol}$

$\text{Mass of He}=n_{He}\times \text{M}\text{.W}\text{of He }$

$\therefore {\text{m}}_{\text{He}}\text{= 0}\text{.0128}\times \text{4 =}0.0512\text{g}$

No. of moles of other gas =$n_{mix}-n_{He}=0.06-0.0128=0.0472$  moles $\text{M}\text{.W of other gas}=5-0.0512=4.948\text{g}$

$\text{No}\text{. of moles }=\frac{\text{mass}}{\text{Gram}\text{.MW}}$

$\Rightarrow 0.0472=\frac{4.948}{\text{M}\text{.W}}$

$\therefore \text{M}\text{.W of other gas}=103.8\text{g}$