JEE Main 2019 — Solutions Question with Solution
From: JEE Main 2019 (Online) 12th April Evening Slot
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)
(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
We know, osmotic pressure () = (Resultant molarity)RT
Resultant molarity =
= = 0.2 mol/lit
Osmotic pressure () = (0.2)0.08206300 = 4.92 atm
Resultant molarity =
= = 0.2 mol/lit
Osmotic pressure () = (0.2)0.08206300 = 4.92 atm
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