Question

A gas X diffuses 1 by root 2 times slower than sulphur dioxide . Find the molecular weight of X

Answer 1

Answer:

Explanation:

(1/2^1/2 )RSO2=Rx

RSO2/Rx=(Mx/MSO2)^1/2

2^1/2=(Mx/64)^1/2                                SO2=32+16*2=64

Mx=64*2=128

Answer 2

Answer:

The molecular mass of gas X will be equal to 128gmol⁻¹.

Explanation:

Given, Rate of diffusion of SO₂ ,$R_{1}=R$

The rate of diffusion of gas X, $R_{2}=\frac{1}{\sqrt{2} } R$

We know, molar mass of $SO_{2}$, $M_{1}=64gmol^{-1}$

By Graham's law: rate of diffusion of a gas is inversely proportional to square root of its molecular weight.

Rate ∝ 1/√M

Therefore  $\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}} }$

$\frac{R}{\frac{1}{\sqrt{2} }R } =\sqrt{\frac{M_{2}}{64 }$

$\sqrt{2}=\sqrt{\frac{M_{2}}{64 }$

On squaring, we get,

$2=\frac{M_{2}}{64}$

$M_{2}=128gmol^{-1}$

Therefore, molar mass of gas X is 128gmol⁻¹.