QJEEM25Cha610: JEE Main 2025 Chemistry - Class 12 JEEM25Cha610 Solution: Arrange the following solutions in order of their increasing…

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

NaCl dissociates (ideally) into two ions:

$\mathrm{NaCl} \;\rightarrow\; \mathrm{Na^+} + \mathrm{Cl^-},$

so $i \approx 2.$

Urea ( $\mathrm{CH_4N_2O}$ ) is a non‐electrolyte (does not dissociate), so $i = 1.$

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

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

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 $\bigl[1 \times 10^{-4}\bigr]$

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

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

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

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

Final Answer

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