Question

An elemental gas has a mass of 10.3 g. If the volume is 58.4 L and the pressure is 101 kPa at a temperature of 2.5 C, what is the gas?

Answer 1

Solving for `"molecular mass"`, we eventually identify the gas as `"helium"`.

Explanation:

If `n=(PV)/(RT)`, `"mass"/"molar mass"=(PV)/(RT)`

And so `"molar mass"=(RTxx"mass")/(PV)`

`=(0.0821*(L*atm)/(K*mol)xx10.3*gxx275.5*K)/(1*atmxx58.4*L)=4.0*g*mol^-1`, which molar mass is consistent with that of helium. (Note that is also consistent with `""^2H_2`.)