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

May 23, 2017

$37 \text{g}$

#### Explanation:

We can use Graham's law of effusion , which relates the relative rates of effusion of gases in terms of their molar masses:

(r_1)/(r_2) = sqrt((MM_2)/(MM_1)

Where
$\frac{{r}_{1}}{{r}_{2}}$ is the ratio of effusion of gas $1$ to gas $2$, and
$M {M}_{1 , 2}$ is the molar mass of the gases $1$ and $2$, respectively.

We're given that the ratio of effusion of gas A to gas B is $0.68$, and that $M {M}_{B}$ is $17 \text{g"/"mol}$, so

$0.68 = \frac{\sqrt{17}}{\sqrt{M {M}_{A}}}$

$M {M}_{A} = {\left(\frac{\sqrt{17}}{0.68}\right)}^{2} = 37 \text{g"/"mol}$

The mass of the gas is thus $37 \text{g}$.