JEE Main 2019 — Solutions Question with Solution
From: JEE Main 2019 (Online) 10th April Morning Slot
Question
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the
vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be.
(molar mass of urea = 60 g mol–1)
Choose an option
Show full solutionCorrect option: B
Correct answer
B0.017 mmHg
Step-by-step explanation
Given that,
wsolute = wurea = 0.6 gm
wsolvent = wH2O = 360 gm
po = 35
We know,
lowering of vapour pressure
p = xsolute po
= po
= 35
=
= 0.017
wsolute = wurea = 0.6 gm
wsolvent = wH2O = 360 gm
po = 35
We know,
lowering of vapour pressure
p = xsolute po
= po
= 35
=
= 0.017
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