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?

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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?

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Answer 1 · 2017-05-23T00:15:51

$37 g$ $37"g"$

Explanation:

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

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M M_{2}}{M M_{1}}}$ $(r_1)/(r_2) = sqrt((MM_2)/(MM_1)$

Where
$\frac{r_{1}}{r_{2}}$ $(r_1)/(r_2)$ is the ratio of effusion of gas $1$ $1$ to gas $2$ $2$ , and
$M M_{1, 2}$ $MM_(1, 2)$ is the molar mass of the gases $1$ $1$ and $2$ $2$ , respectively.

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

$0.68 = \frac{\sqrt{17}}{\sqrt{M M_{A}}}$ $0.68 = (sqrt17)/sqrt(MM_A)$

$M M_{A} = {(\frac{\sqrt{17}}{0.68})}^{2} = 37 \frac{g}{mol}$ $MM_A = ((sqrt(17))/(0.68))^2 = 37"g"/"mol"$

The mass of the gas is thus $37 g$ $37"g"$ .

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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?

1 Answer

Explanation:

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