Verified Solution • 2025–26 Session moderate

QJEEM21Cha9ac: JEE Main 2021 Chemistry - Class 12 JEEM21Cha9ac Solution: 1.46 g of a biopolymer dissolved in a 100 mL water at 300 K…

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 104 g mol −- 1. (Round off to the Nearest Integer)

[Use : R = 0.083 L bar mol −- 1 K −- 1]

Step-by-Step Verified Solution

π\pi = CRT;

π\pi = osmotic pressure

C = molarity

T = Temperature of solution

let the molar mass be M gm/mol

2.42 ×\times 10 −- 3 bar = (1.46gMgm/mol)0.1l×(0.083l−barmol−K)×(300K){{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)

⇒\Rightarrow M = 15.02 ×\times 104 g/mol

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