0.30 g of gas was found to occupy a volume of 82.0 mL at `27^(@)C` and 3 atm. Pressure. The molecular mass of the gas is

Options

A. `60`
B. `30`
C. `90`
D. unpredictable

Answer

B

Explanation

`PV = nRT " or " PV = (w)/(M) RT`
`:. M = (wRT)/(PV)`
`= ((0.30 g) xx (0.0820 L atm K^(-1) mol^(-1)) xx (300 K))/((3 atm) xx (82.0 xx 10^(-3) L)) = 30`