Question

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?

Answer 1

`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"`.