JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Solutions

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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

g mol^{- 1}

H = 1

C = 12

N = 14

O = 16

Cl = 35.5

Ca = 40

Al = 27

and

S = 32

]

(A)I $<$ II $<$ III $<$ IV✓ Correct
(B)III $<$ I $<$ II $<$ IV
(C)II $<$ III $<$ I $<$ IV
(D)II $<$ III $<$ IV $<$ I

Answer: (A)

Solution

Boiling-point elevation follows

Δ T_{b} \propto i m

, so the solutions are ranked by the product of van't Hoff factor and molar concentration. Glucose gives

\frac{2.2/180}{0.125} = 0.098 M

with

i = 1

, product

0.098

CaC l_{2}

gives

\frac{1.9/111}{0.25} = 0.068 M

with

i = 3

, product

0.205

. Urea gives

\frac{9.0/60}{0.5} = 0.30 M

with

i = 1

, product

0.30

A l_{2} (S O_{4})_{3}

gives

\frac{20.5/342}{0.75} = 0.080 M

with

i = 5

, product

0.40

. The increasing order of boiling point is glucose

<

CaC l_{2}

<

urea

<

A l_{2} (S O_{4})_{3}

, that is I

<

<

III

<

IV.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Solutions

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