2.8 g of a gas at 1 atm and 273K occupies a volume of 2.24 litres, the gas can not be:

Let the molar mass of the gas be M gm/mole

Applying Ideal gas equation to given situation

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ 1\times 2.24=\frac{2.8}{\mathrm{M}}\times 0.0821\times 273 \end{gathered}$$

M = 28 gm/mole

So, the gas can't be Oxygen because its molar mass is 32 gm/mole

Hence, Option "A" is the correct answer