1 . 46g of a biopolymer dissolved in a 100mL water at 300K exerted...

$\pi = CRT; \pi =$ osmotic pressure
$C =$ molarity
$T =$ Temperature of solution let the molar mass be

Molar mass of polymer = $Mgm / mol$

$2 . 42 \times {10}^{- 3}bar = = \dfrac{\left(\dfrac{1 . 46g}{Mgm / mol}\right)}{0 . 1ℓ} \times \left(\dfrac{0 . 083ℓ . bar}{mol . K}\right) \times \left(300K\right)$

$⇒ M = 15 . 02 \times {10}^{4}g / mol$