van der Waals' gas equation for $\mu$ mole of real gas
$ \, \, \, \, \, \, \, \, \bigg( p + \frac{\mu^2 a}{V^2}\bigg) (V - \mu b) = \mu \, RT$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, p = \bigg( \frac{\mu RT}{V - \mu b}\bigg) - \frac{\mu^2 a}{V^2}$
Given equation, $ \, \, \, \, \, \, \, \, \, \, p = \bigg( \frac{RT}{2V - b} - \frac{a}{4b^2}\bigg)$
On comparing the given equation with this standard equation, we get
$ \, \, \, \, \, \, \, \mu = \frac{1}{2}$
Hence, $\mu = \frac{m}{M} \Rightarrow $ mass of gas
$ \, \, \, \, \, \, \, \, \, \, \, m = \mu M = \frac{1}{2} \times 44 = 22g$