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)} $