Arrange the following solutions in order of their increasing boiling points. (i) $10^{-4} \mathrm{M} \mathrm{NaCl}$ (ii) $10^{-4} \mathrm{M}$ Urea (iii) $10^{-3} \mathrm{M} \mathrm{NaCl}$ (iv) $10^{-2} \mathrm{M} \mathrm{NaCl}$

(B) (ii) < (i) ≅ (iii) < (iv)

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Solutions

Arrange the following solutions in order of their increasing boiling points.
(i)

1 0^{- 4} M NaCl

(ii)

1 0^{- 4} M

Urea
(iii)

1 0^{- 3} M NaCl

(iv)

1 0^{- 2} M NaCl

(A)(ii) < (i) < (iii) < (iv)✓ Correct
(B)(ii) < (i) ≅ (iii) < (iv)
(C)(i) < (ii) < (iii) < (iv)
(D)(iv) < (iii) < (i) < (ii)

Answer: (A)

Solution

Boiling point elevation is a colligative property,

Δ T_{b} = i K_{b} m \propto i \cdot C

, where

i

is the van't Hoff factor.

NaCl dissociates into two ions, so

i = 2

; urea does not dissociate, so

i = 1

i \cdot C

values: urea

1 0^{- 4} M (1 \times 1 0^{- 4})

< NaCl

1 0^{- 4} M (2 \times 1 0^{- 4})

< NaCl

1 0^{- 3} M (2 \times 1 0^{- 3})

< NaCl

1 0^{- 2} M (2 \times 1 0^{- 2})

Boiling point rises with

i \cdot C

, giving the increasing order (ii) < (i) < (iii) < (iv).

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option

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