Question

Helium is found to diffuse four times more rapidly than an unknown gas. What is the approximate molar mass of the unknown gas?

Answer 1

Around `"64 g/mol"`.

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You can start from any expression for the speed. The RMS speed is a fine choice:

`v_(RMS) = sqrt((3RT)/M)`,

where `R` and `T` are from the ideal gas law, and `M` is the molar mass of the gas in `"kg/mol"`.

The ratio of two speeds `v` is directly proportional to the ratio of the gas diffusion rates `z` (the `3RT` cancels out):

`(v_(RMS,B))/(v_(RMS,A)) = color(green)(z_B/z_A = sqrt(M_A/M_B))`

which is Graham's law of diffusion; the gas with more mass per particle diffuses more slowly. (In this case, the molar mass can now be in `"g/mol"` and it won't matter.)

Since helium is known to diffuse `4` times as fast as unknown gas `B`, then off the top of my head, helium is probably `4^2 = 16` times as light (one-sixteenth the molar mass of `B`).

Let's check mathematically.

`z_B/z_A = sqrt(M_A/M_B)`

If we assign helium as `A`, then the unknown gas is `B` and `z_B/z_A = 1/4`.

`=> 1/4 = sqrt(4.0026/M_B)`

`=> 1/16 = 4.0026/M_B`

`=> color(blue)(M_B) = 4.0026 xx 16 ~~ color(blue)("64 g/mol")`

So the unknown gas has about `bb16` times higher of a molar mass.