1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42×10^(-3)bar The molar mass of the biopolymer is -×10^4 g mol^(-1). (Round off to the Nearest Integer) [Use: R=0.083 L bar mol^(-1) K^(-1)

Q24 / 90 · JEE Main 2021 (27 Jul Shift 1) · Chemistry

$1.46 g$ of a biopolymer dissolved in a $100 mL$ water at $300 K$ exerted an osmotic pressure of $2.42 \times 10^{- 3} bar$
The molar mass of the biopolymer is $- \times 10^{4} g$ ${mol}^{- 1} .$ (Round off to the Nearest Integer)
[Use: $R = 0.083 L$ bar ${mol}^{- 1} K^{- 1}]$

Correct answer: 15

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

Molar mass of polymer = $M gm / mol$

$2.42 \times 10^{- 3} bar = = \frac{(\frac{1.46 g}{M gm / mol})}{0.1 ℓ} \times (\frac{0.083 ℓ . bar}{mol . K}) \times (300 K)$

$\Rightarrow M = 15.02 \times 10^{4} g / mol$

Hence, the correct answer is 15.