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
May 23, 2017

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:

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

Where
(r_1)/(r_2) is the ratio of effusion of gas 1 to gas 2, and
MM_(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 MM_B is 17"g"/"mol", so

0.68 = (sqrt17)/sqrt(MM_A)

MM_A = ((sqrt(17))/(0.68))^2 = 37"g"/"mol"

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