JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Consider the following aqueous solutions.

I. 2.2 g Glucose in 125 mL of solution.

II. 1.9 g Calcium chloride in 250 mL of solution.

III. 9.0 g Urea in 500 mL of solution.

IV. 20.5 g Aluminium sulphate in 750 mL of solution.

The correct increasing order of boiling point of these solutions will be :

[Given : Molar mass in $\mathrm{g} \mathrm{mol}^{-1}: \mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Cl}=35.5, \mathrm{Ca}=40, \mathrm{Al}=27$ and $\mathrm{S}=32$ ]

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\text { III }<\text { I }<\text { II }<\text { IV }$
B
$\mathrm{I}<\mathrm{II}<\mathrm{III}<\mathrm{IV}$
C
$\text { II }<\text { III }<\text { I }<\text { IV }$
D
$\text { II }<\text { III }<\text { IV }<\text { I }$

Did you get this right?

Solution

$\Delta T_b = i \cdot k_b \cdot m$

For dilute solution ( $M = m$ )

Molarity	$i \times m$
(I) $M_{glucose} = \dfrac{2.2}{180} \times \dfrac{1000}{125} = 0.098$	$0.098 \times 1$
(II) $M_{CaCl_2} = \dfrac{1.9}{111} \times \dfrac{1000}{250} = 0.068$	$0.068 \times 3$
(III) $M_{urea} = \dfrac{9}{60} \times \dfrac{1000}{500} = 0.3$	$0.3 \times 1$
(IV) $M_{Al_2(SO_4)_2} = \dfrac{20.5}{342} \times \dfrac{1000}{750} \approx 0.08$	$0.08 \times 5$

Order of $\Delta T_b$ = $Al_2(SO_4)_3$ > Urea > $CaCl_2$ > Glucose

So order of BP = $Al_2(SO_4)_3$ > Urea > $CaCl_2$ > Glucose

So Answer will be I < II < III < IV

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 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

Options

$\text { III }<\text { I }<\text { II }<\text { IV }$

$\mathrm{I}<\mathrm{II}<\mathrm{III}<\mathrm{IV}$

$\text { II }<\text { III }<\text { I }<\text { IV }$

$\text { II }<\text { III }<\text { IV }<\text { I }$

Solution

$\Delta T_b = i \cdot k_b \cdot m$

For dilute solution ( $M = m$ )

Molarity	$i \times m$
(I) $M_{glucose} = \dfrac{2.2}{180} \times \dfrac{1000}{125} = 0.098$	$0.098 \times 1$
(II) $M_{CaCl_2} = \dfrac{1.9}{111} \times \dfrac{1000}{250} = 0.068$	$0.068 \times 3$
(III) $M_{urea} = \dfrac{9}{60} \times \dfrac{1000}{500} = 0.3$	$0.3 \times 1$
(IV) $M_{Al_2(SO_4)_2} = \dfrac{20.5}{342} \times \dfrac{1000}{750} \approx 0.08$	$0.08 \times 5$

Order of $\Delta T_b$ = $Al_2(SO_4)_3$ > Urea > $CaCl_2$ > Glucose

So order of BP = $Al_2(SO_4)_3$ > Urea > $CaCl_2$ > Glucose

So Answer will be I < II < III < IV

JEE Main 2026 — 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 2026 — 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