JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

Consider the following aqueous solutions.

I. 2.2 g Glucose in 125 mL of solution.

II. 1.9 g Calcium chloride in 250 mL of solution.

III. 9.0 g Urea in 500 mL of solution.

IV. 20.5 g Aluminium sulphate in 750 mL of solution.

The correct increasing order of boiling point of these solutions will be:

[Given: Molar mass in g mol $^{- 1}$ : H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32]

Solution

Boiling point elevation $Δ T_{b} = i \cdot K_{b} \cdot m$ . For dilute solutions take molality $\approx$ molarity. Compute $i \times M$ for each:

I. Glucose ( $i = 1$ ): $M = \frac{2.2}{180} \times \frac{1000}{125} \approx 0.098$ . $i M = 0.098$ .

II. CaCl $_{2}$ ( $i = 3$ , $M_{molar} = 111$ ): $M = \frac{1.9}{111} \times \frac{1000}{250} \approx 0.068$ . $i M \approx 0.205$ .

III. Urea ( $i = 1$ , $M_{molar} = 60$ ): $M = \frac{9}{60} \times \frac{1000}{500} = 0.30$ . $i M = 0.30$ .

IV. Al $_{2}$ (SO $_{4}$ ) $_{3}$ ( $i = 5$ , $M_{molar} = 342$ ): $M = \frac{20.5}{342} \times \frac{1000}{750} \approx 0.08$ . $i M \approx 0.40$ .

Order: $0.098 < 0.205 < 0.30 < 0.40$ , i.e. I $<$ II $<$ III $<$ IV.

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