Question

In an experiment, an unknown gas effuses at one-half the speed of oxygen gas, which has a molar mass of 32 g/mol. What might the unknown gas be?

Answer 1

It could be iodine..

Explanation:

We can solve this equation using Graham's law of effusion:

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

We can substitute gas `"X"` and `"oxygen"` in for 1 and 2, respectively, to get

`(r_"X")/(r_ ("O"_2)) = sqrt((MM_ ("O"_2))/(MM_"X"))`

Since `"O"_2` effuses at a rate twice that of gas `"X"`, the ratio of the rate of effusion of `"X"` to `"O"_2` is `0.5`:

`0.5 = sqrt((MM_ ("O"_2))/(MM_"X"))`

Since the molar mass of `"O"_2` is given as `32.00` `"g/mol"`, we then have

`0.5 = sqrt((32color(white)(l)"g/mol")/(MM_"X"))`

Using algebra to solve for `MM_"X"`:

`0.5 = sqrt(32color(white)(l)"g/mol")/(sqrt(MM_"X"))`

`sqrt(MM_"X") = sqrt(32color(white)(l)"g/mol")/0.5`

`MM_"X" = (sqrt(32color(white)(l)"g/mol")/0.5)^2 = color(red)(128` `color(red)("g/mol"`

The molar mass of the unknown gas is thus `128` `"g/mol"`

A gas close to this value is `sfcolor(blue)("iodine"`, with molar mass `126.90` `"g/mol"`.