Chapter 6, Problem 6.169AP

Interpretation Introduction

Interpretation: A sample of ${\text{N}}_2$ is given to require $240\text{s}$ to diffuse through a porous plug. It takes $530\text{s}$ for an equal number of moles of an unknown gas $\text{X}$ through the plug under the same conditions of temperature and pressure. The molar mass of gas $\text{X}$ is to be calculated.

Concept introduction: The process in which the gas is escapes through a tiny hole from its container into the region where pressure is low is called as effusion. Graham’s gives a relationship that the rate of effusion of gases is inversely proportional to the square root of their molar masses. This relationship is known as Graham’s law of effusion.

To determine: The molar mass of unknown gas $\text{X}$.