Chapter 3, Problem 3D.7E

Interpretation Introduction

Interpretation:

Molar mass and molecular formula of compound with empirical formula ${\text{C}}_2{\text{H}}_3$ has to be determined.

Concept Introduction:

Subscript of empirical formula is multiplied by value of $n$ to give molecular formula. The value of $n$ can be calculated through division of molecular mass by empirical formula mass as follows:

  $n=\frac{\text{molecular mass}}{\text{empirical formula mass}}$

Effusion is explained as movement of the gas molecule through small hole.

Diffusion can be explained as when 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 $\text{A}$.