# Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g/mol, what is the molar mass of Gas A?

Jul 10, 2017

$\text{molar mass of A} = 37$ $\text{g/mol}$

#### Explanation:

We can use Graham's law of effusion to solve this:

$\frac{{r}_{A}}{{r}_{B}} = \sqrt{\frac{{M}_{B}}{{M}_{A}}}$

where

• $\frac{{r}_{A}}{{r}_{B}}$ is a number representing the ratio of effusion rates of gas A to gas B, in this case $0.68 : 1 = 0.68$

• ${M}_{A}$ and ${M}_{2}$ are the molar masses of gases A and B, respectively.

We know the molar mass of gas B is $17$ $\text{g/mol}$, so let's solve this for ${M}_{A}$:

$0.68 = \frac{\sqrt{17 \textcolor{w h i t e}{l} \text{g/mol}}}{\sqrt{{M}_{B}}}$

$\sqrt{{M}_{B}} = \frac{\sqrt{17 \textcolor{w h i t e}{l} \text{g/mol}}}{0.68}$

M_B = (sqrt(17color(white)(l)"g/mol")/0.68)^2 = color(red)(37 color(red)("g/mol"

rounded to $2$ significant figures.