$1.46 \mathrm{~g}$ of a biopolymer dissolved in a $100 \mathrm{~mL}$ water at $300 \mathrm{~K}$ exerted an osmotic pressure of $2.42 \times 10^{-3}$ bar.
The molar mass of the biopolymer is_________ $\times 10^{4} \mathrm{~g}$ $\mathrm{mol}^{-1}$. (Round off to the Nearest Integer)
$\mathrm{mol}^{-1}$. (Round off to the Nearest Integer)
[Use : $\mathrm{R}=0.083 \mathrm{~L}$ bar $\mathrm{mol}^{-1} \mathrm{~K}^{-1}$ ]
$\pi=$ CRT $; \pi=$ osmotic pressure
$\mathrm{C}=$ molarity
$\mathrm{T}=$ Temperature of solution
let the molar mass be $\mathrm{M} \mathrm{gm} / \mathrm{mol}$
$2.42 \times 10^{-3}$ bar $=$
$\frac{\left(\frac{1.46 \mathrm{~g}}{\mathrm{Mgm} / \mathrm{mol}}\right)}{0.1 \ell} \times\left(\frac{0.083 \ell-\mathrm{bar}}{\mathrm{mol}-\mathrm{K}}\right) \times(300 \mathrm{~K})$
$\Rightarrow \mathrm{M}=15.02 \times 10^{4} \mathrm{~g} / \mathrm{mol}$