Solutions Part 2

Concept:The boiling point of a solution increases with the number of solute particles.This is given by $\Delta T_b = i \cdot k_b \cdot m$, where $i$ is van’t Hoff factor and $m$ is molality.For dilute aqueous solutions, molality $m$ is nearly equal to molarity $M$ (since $k_b$ of water is constant).Explanation:Calculate molarity ($M$) for each solution and multiply by $i$.Use molar masses: Glucose ($\mathrm{C_6H_{12}O_6}$) = 180 g/mol, $\mathrm{CaCl_2}$ = 111 g/mol, Urea ($\mathrm{CH_4N_2O}$) = 60 g/mol, $\mathrm{Al_2(SO_4)_3}$ = 342 g/mol.I. Glucose (non‑electrolyte, $i = 1$): $M = \frac{2.2}{180} \times \frac{1000}{125} = 0.098$; $i \times M = 0.098$.II. $\mathrm{CaCl_2}$ ($i = 3$): $M = \frac{1.9}{111} \times \frac{1000}{250} = 0.068$; $i \times M = 0.068 \times 3 = 0.204$.III. Urea (non‑electrolyte, $i = 1$): $M = \frac{9}{60} \times \frac{1000}{500} = 0.3$; $i \times M = 0.3$.IV. $\mathrm{Al_2(SO_4)_3}$ ($i = 5$): $M = \frac{20.5}{342} \times \frac{1000}{750} \approx 0.08$; $i \times M \approx 0.08 \times 5 = 0.4$.Order of $i \times M$: $0.098 < 0.204 < 0.3 < 0.4$.Thus, increasing boiling point order is: I < II < III < IV.Shortcut:For dilute solutions, directly compare $i \times M$ values. Non‑electrolytes have $i = 1$; electrolytes dissociate completely (given).Answer:Option D: I $<$ II $<$ III $<$ IV