Question

What is the molar mass of a .250 gas that occupies 1.01 L at 31°C and 1.44 atm?

Answer 1

Approx. `4.3*g*mol^-1`.

Explanation:

We assume Ideal Gas Behaviour, by which we can calculate a molar mass:

`"Mass"/"Molar Mass"=(PV)/(RT)`

And thus `"Molar Mass"` `=` `(RT)/(PV)xx"Mass"`

`=` `(0.0821*L*atm*K^-1*mol^-1xx304*Kxx0.250*g)/(1.44*atmxx1.01*L)`

`=` `??`

What do you think is the identity of the gas?

And why did I go to the bother of writing out the individual units, pounds, ounces, furlongs, in the calculation? It was to help save me from making an error. I wanted an answer in `g*mol^-1`, and when I cancelled out the common terms, I got such an answer. This persuades me that I am not entirely clueless (only partially clueless today!).