In the first case,
π1 = C1RT
i.e., 4.98 bar = \(\frac{36g\times R\times300K}{180g\,mol^{-1}}\) ——– (1)
In second case,
π2 = C2RT
i.e., 1.52 bar = C2R × 300 K ———- (2)
\(\frac{Eqn.(2)}{Eqn.(1)}\Rightarrow \frac{1.52\,bar}{4.98\,bar}\)\(=\frac{C_2\times R\times 300K\times 180\,g\,mol^{-1}}{36\,gL^{-1}\times R\,\times\,300K}\)
∴ \(C_2=\frac{1.52\,bar\times\,36\,g\,L^{-1}}{4.98\,bar\times 180\,g\,mol^{-1}}\)
= 0.061 mol L-1