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