[ઉપયોગ : ${R}=0.083\, {~L}\, bar \,{mol}^{-1} \,{~K}^{-1}$ ]
\(; \pi=\) osmotic pressure
\({C}=\) molarity
\(T=\) Temperature of solution
let the molar mass be \({M}\, {gm} / {mol}\)
\(2.42 \times 10^{-3}\, {bar}=\frac{\left(\frac{1.46\, {~g}}{{M\,gm} / {mol}}\right)}{0.1\, \ell} \times\left(\frac{0.083\, \ell-{bar}}{{mol}-{K}}\right) \times(300\, {~K})\)
\(\Rightarrow \quad {M}=15.02 \times 10^{4}\, {~g} / {mol}\)