Question

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?

Answer 1

`"molar mass of A" = 37` `"g/mol"`

Explanation:

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

`(r_A)/(r_B) = sqrt((M_B)/(M_A))`

where

• `(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` `"g/mol"`, so let's solve this for `M_A`:

`0.68 = sqrt(17color(white)(l)"g/mol")/(sqrt(M_B))`

`sqrt(M_B) = sqrt(17color(white)(l)"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.