Chapter 12, Problem 44A

Interpretation Introduction

Interpretation:

The molar mass of gas with respect to diffusion of $O_2$ needs to be determined.

Concept introduction:

Rate of diffusion is inversely proportional to square root of molar mass of gas as follows:

$R\propto \frac{1}{\sqrt{M}}$

Where, R is rate of diffusion & M is molar mass.

Answer to Problem 44A

The molar mass of gas is $3.55gm/mole$.

Explanation of Solution

Rate of diffusion given by formula

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$

Here $O_2$ and unknown gas is used.

$\frac{R_1}{R_2}=\sqrt{\frac{M_{Gas}}{M_{O_2}}}\Rightarrow \frac{R_{O_2}}{3R_{O_2}}=\sqrt{\frac{M_G}{32}}\Rightarrow where{R}_G=3R_{O_2}$

$\frac{M_G}{32}={\left(\frac{1}{3}\right)}^2\Rightarrow \frac{M_G}{32}=\frac{1}{9}\Rightarrow M_G=\frac{32}{9}=3.55gm/mole$

On the basis of Graham’s Law, mass of gas is calculated by Graham’s formula.