Chapter 3, Problem 3D.1AST

Interpretation Introduction

Interpretation:

Molar mass of volatile compound has to be determined.

Concept Introduction:

Effusion is explained as movement of the gas molecule through small hole. Diffusion can be explained as one gas molecule mix with another gas molecule by random motion.

Graham’s law of effusion gives relation of rate of effusion with molar mass. Rate of effusion and square root of molar mass are inversely related. The mathematical expression is as follows:

  $\frac{\text{Effusion rate of A}}{\text{Effusion rate of B}}=\sqrt{\frac{M_{\text{B}}}{M_{\text{A}}}}$

Here,

$M_{\text{B}}$ is molar mass of gas $\text{B}$.

$M_{\text{A}}$ is molar mass of gas $\text{A}$.