JEE Main 2021 — Solutions Question with Solution
From: JEE Main 2021 (Online) 27th July Morning Shift
Question
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 103 bar.
The molar mass of the biopolymer is _____________ 104 g mol1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol1 K1]
The molar mass of the biopolymer is _____________ 104 g mol1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol1 K1]
Enter your answer
Show full solutionCorrect answer: 15.02
Correct answer
15.02
Step-by-step explanation
= CRT;
= osmotic pressure
C = molarity
T = Temperature of solution
let the molar mass be M gm/mol
2.42 103 bar =
M = 15.02 104 g/mol
= osmotic pressure
C = molarity
T = Temperature of solution
let the molar mass be M gm/mol
2.42 103 bar =
M = 15.02 104 g/mol
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