JEE Main 2019 — Chemistry Physical Chemistry

2019mcqmedium

At room temperature, a dilute solution of urea is prepared by dissolving $0.60 g$ 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 $=60 g mo{l}^{-1}$ )

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

A
$0.028 mm Hg$
B
$0.027 mm Hg$
C
$0.031 mm Hg$
D
$0.017 mm Hg$

Did you get this right?

Solution

Relative lowering in vapour Pressure

$\Rightarrow$ Mole of urea $({n}_{B})=\frac{0.6}{60}={10}^{-2} mole$ ;

Mole of water $({n}_{A})=\frac{360}{18}=20$

$\frac{{P}^{o}-{P}_{s}}{{P}^{o}}={x}_{B}=\frac{{P}^{o}-{P}_{s}}{{P}_{o}}=\frac{{n}_{B}}{{n}_{n}+{n}_{B}}$

Here $={P}^{o}=V.P.$ of pure solvent

${P}_{s}=V.P$ of solution.

${n}_{A}+{n}_{B}\simeq {n}_{A}$

$\frac{{P}^{o}-{P}_{s}}{{P}^{o}}=\frac{{n}_{B}}{{n}_{A}}$

$lowering of vapour pressure({P}^{o}-{P}_{s})=\frac{{n}_{B}}{{n}_{A}}\times {P}^{o}$

${P}^{o}-{P}_{s}=\frac{{10}^{-2}}{20}\times 35$

${P}^{o}-{P}_{s}=0.0175 mm$ of $Hg$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2019 — Chemistry Physical Chemistry

2019mcqmedium

(molar mass of urea $=60 g mo{l}^{-1}$ )

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

A
$0.028 mm Hg$
B
$0.027 mm Hg$
C
$0.031 mm Hg$
D
$0.017 mm Hg$

Did you get this right?

Solution

Relative lowering in vapour Pressure

$\Rightarrow$ Mole of urea $({n}_{B})=\frac{0.6}{60}={10}^{-2} mole$ ;

Mole of water $({n}_{A})=\frac{360}{18}=20$

$\frac{{P}^{o}-{P}_{s}}{{P}^{o}}={x}_{B}=\frac{{P}^{o}-{P}_{s}}{{P}_{o}}=\frac{{n}_{B}}{{n}_{n}+{n}_{B}}$

Here $={P}^{o}=V.P.$ of pure solvent

${P}_{s}=V.P$ of solution.

${n}_{A}+{n}_{B}\simeq {n}_{A}$

$\frac{{P}^{o}-{P}_{s}}{{P}^{o}}=\frac{{n}_{B}}{{n}_{A}}$

$lowering of vapour pressure({P}^{o}-{P}_{s})=\frac{{n}_{B}}{{n}_{A}}\times {P}^{o}$

${P}^{o}-{P}_{s}=\frac{{10}^{-2}}{20}\times 35$

${P}^{o}-{P}_{s}=0.0175 mm$ of $Hg$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry