Verified Solution • 2025–26 Session moderate

QJEEM19Ch7743: JEE Main 2019 Chemistry - Class 12 JEEM19Ch7743 Solution: A solution is prepared by dissolving 0.6 g of urea (molar mass…

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27oC. The osmotic pressure of the solution is :
(R = 0.08206 L atm K–1 mol–1)

Multiple Choice Options

  • A 8.2 atm
  • B 2.46 atm
  • C 4.92 atm
  • D 1.64 atm

Step-by-Step Verified Solution

We know, osmotic pressure ( π\pi ) = (Resultant molarity) ×\times R ×\times T

Resultant molarity = Total moles of all solutionsTotal volume(in lit){{Total\,moles\,of\,all\,solutions} \over {Total\,volume(in\,lit)}}
= 0.660+1.81800.1{{{{0.6} \over {60}} + {{1.8} \over {180}}} \over {0.1}} = 0.2 mol/lit

∴\therefore Osmotic pressure ( π\pi ) = (0.2) ×\times 0.08206 ×\times 300 = 4.92 atm

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring