Chapter 13, Problem 13.56P

Interpretation Introduction

Interpretation:

Volume percentage of $\text{CO}$ has to be calculated.

Concept Introduction:

Graham’s law suggests that the rate of diffusion or effusion of gases varies inversely to the square root of its molar mass. Mathematically it can also be stated as,

  $\frac{t_{\text{B}}}{t_{\text{A}}}={\left(\frac{M_{\text{B}}}{M_{\text{A}}}\right)}^{1/2}$

Here

$t_{\text{B}}$ representeffusion time taken by of gas B.

$t_{\text{A}}$ representeffusion time taken of gas A.

${\text{Rate}}_{\text{A}}$ represent rate of effusion of gas A.

${\text{Rate}}_{\text{B}}$ represent rate of effusion of gas B.

The expression to evaluate molecular weight of mixture is given as follows:

  $\text{Molecular weight of mixture}=\left[\begin{array}{l}\left(\text{volume fraction of A}\right)\left(\text{Molar mass of A}\right)+\\ \left(\text{volume fraction of B}\right)\left(\text{Molar mass of B}\right)\end{array}\right]$

Sum of all volume fractions is unity.