JEE Main 2019ChemistrySolutionsMediumMCQ

JEE Main 2019 — Solutions Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

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)

Choose an option

Show full solutionCorrect option: C
Correct answer
C4.92 atm

Step-by-step explanation

i factor of glucose and urea are 1 : 1 respectively.

Moles of glucose =weightMolecular wieght=1.8180=10-2

Moles of urea =0.660=10-2

Total mole of solute =10-2+10-2=2×10-2

Concentration of solution =2×10-2100×10-3=0.2

Osmotic pressure (π)=CRT

=0.2×0.0821×300=4.926 atm

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About this question

This is a previous-year question from JEE Main 2019, covering the Solutions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.