Question

Question: A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.

Short answer

The molecular mass of the unknown gas is $65.97\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}.$

Step-by-step solution

Concept Introduction

Graham's law states that a gas's rate of diffusion or effusion is inversely related to the square root of its molecular weight.

Information Provided

• The rate of diffusion of the unknown gas$=83.3\mathrm{m}\mathrm{L}/\mathrm{s}$
• The rate of diffusion of$\mathrm{C}{\mathrm{O}}_2=102\mathrm{m}\mathrm{L}/\mathrm{s}$
• The molar mass of $\mathrm{C}{\mathrm{O}}_2=44\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases $A$ and $B$ are at the same temperature and pressure, then the relative rates of diffusion for unknown gas and $\mathrm{C}{\mathrm{O}}_2$ will be:

Relative rates of diffusion for unknown gas and$\mathrm{C}{\mathrm{O}}_2$–

$\begin{array}{l}\frac{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}\mathrm{g}\mathrm{a}\mathrm{s}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{u}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{C}{\mathrm{O}}_2}=\frac{\sqrt{{\mathrm{M}}_{\mathrm{C}{\mathrm{O}}_2}}}{\sqrt{{\mathrm{M}}_{\mathrm{u}\mathrm{n}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}\mathrm{g}\mathrm{a}\mathrm{s}}}},whereMisthemolarmass.\end{array}$

Rearrange the equation and solve as shown below:

$\begin{array}{c}\frac{83.3\mathrm{m}\mathrm{L}/\mathrm{s}}{102\mathrm{m}\mathrm{L}/\mathrm{s}}=\sqrt{\frac{14\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}}{{\mathrm{M}}_{\mathrm{u}\mathrm{n}\mathrm{k}\mathrm{u}\mathrm{r}\mathrm{o}\mathrm{w}\mathrm{n}\mathrm{g}\mathrm{a}\mathrm{s}}}}\\ {\mathrm{M}}_{\mathrm{u}\mathrm{n}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}\mathrm{g}\mathrm{a}\mathrm{s}}={\left(\frac{102\mathrm{m}\mathrm{L}/\mathrm{s}}{83.3\mathrm{m}\mathrm{L}/\mathrm{s}}\right)}^2\times 44\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}\\ =65.97\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}.\end{array}$

Therefore, the result obtained is $65.97\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}.$