Verified Solution • 2025–26 Session moderate

QJEEM19Ch8979: JEE Main 2019 Chemistry - Class 12 JEEM19Ch8979 Solution: At room temperature, a dilute solution of urea is prepared by…

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)

Multiple Choice Options

A 0.031 mmHg
B 0.017 mmHg
C 0.028 mmHg
D 0.027 mmHg

Step-by-Step Verified Solution

Given that,

w_solute = w_urea = 0.6 gm

w_solvent = w_H₂O = 360 gm

p^o = 35

We know,

lowering of vapour pressure

Δ

p = x_solute

\times

p^o

=

\frac{n_{u r e a}}{n_{u r e a} + n_{H_{2} O}}

\times

p^o

=

\frac{\frac{0.6}{60}}{\frac{0.6}{60} + \frac{360}{18}}

\times

35

=

\frac{10^{- 2}}{20} \times 35

= 0.017

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