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

Arrange the solutions in order of increasing boiling points.

We recall that boiling point elevation depends on the total solute particle concentration.

$$\Delta T_b = i \cdot K_b \cdot m$$

Here, $$i$$ is the van't Hoff factor.

Next, we calculate the effective concentration for each solution.

For (i) $$10^{-4}$$ M NaCl, which dissociates as NaCl → Na⁺ + Cl⁻, the van't Hoff factor is $$i = 2$$ giving an effective concentration of $$2 \times 10^{-4}$$ M.

For (ii) $$10^{-4}$$ M urea, a non-electrolyte with $$i = 1$$, the effective concentration is $$10^{-4}$$ M.

For (iii) $$10^{-3}$$ M NaCl with $$i = 2$$, the effective concentration becomes $$2 \times 10^{-3}$$ M.

For (iv) $$10^{-2}$$ M NaCl with $$i = 2$$, the effective concentration becomes $$2 \times 10^{-2}$$ M.

Ordering these effective concentrations yields

$$10^{-4} < 2 \times 10^{-4} < 2 \times 10^{-3} < 2 \times 10^{-2}$$

Hence, in terms of the given options:

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

Therefore, the solutions in order of increasing boiling point are (ii) < (i) < (iii) < (iv).