JEE Main 2025 — Solutions Question with Solution

Step 1: Identify the van’t Hoff factor ($i$) for each solute

NaCl dissociates (ideally) into two ions:

$\mathrm{NaCl}\to \mathrm{N}{\mathrm{a}}^{+}+\mathrm{C}{\mathrm{l}}^{-},$

so $i\approx 2.$

Urea ($\mathrm{C}{\mathrm{H}}_4{\mathrm{N}}_2\mathrm{O}$) is a non‐electrolyte (does not dissociate), so $i=1.$

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Step 2: Effective molar concentration of particles

The total particle concentration for each solution is approximately $(i\times \text{molarity})$.

(i) $10^{-4}M$ NaCl

$\text{Effective concentration}=2\times 10^{-4}=2\times 10^{-4}.$

(ii) $10^{-4}M$ Urea

$\text{Effective concentration}=1\times 10^{-4}=1\times 10^{-4}.$

(iii) $10^{-3}M$ NaCl

$\text{Effective concentration}=2\times 10^{-3}=2\times 10^{-3}.$

(iv) $10^{-2}M$ NaCl

$\text{Effective concentration}=2\times 10^{-2}=2\times 10^{-2}.$

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Step 3: Compare to rank the boiling points

A larger total particle concentration (and hence larger colligative effect) corresponds to a higher boiling point. Arrange from lowest to highest:

Lowest: $10^{-4}M$ Urea $[1\times 10^{-4}]$

Next: $10^{-4}M$ NaCl $[2\times 10^{-4}]$

Next: $10^{-3}M$ NaCl $[2\times 10^{-3}]$

Highest: $10^{-2}M$ NaCl $[2\times 10^{-2}]$

Hence, in the format $(\text{ii})<(\text{i})<(\text{iii})<(\text{iv})$.

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Final Answer

$\boxed{\text{(ii) }<\text{(i) }<\text{(iii) }<\text{(iv)}}\text{(Option B)}$