Q. 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]

Correct Answer: I < II < III < IV

Explanation

Elevation in boiling point depends on the quantity i × m, where i = van’t Hoff factor and m = molality.

---

Step 1: Calculate molarity (approx. molality)

Solution	Moles	Volume (L)	Molarity (M)
I (Glucose)	2.2 / 180 = 0.0122	0.125	0.0976
II (CaCl2)	1.9 / 111 = 0.0171	0.25	0.0684
III (Urea)	9 / 60 = 0.15	0.5	0.30
IV (Al2(SO4)3)	20.5 / 342 = 0.06	0.75	0.08

---

Step 2: Apply van’t Hoff factor

Glucose → non-electrolyte → i = 1

Urea → non-electrolyte → i = 1

CaCl₂ → Ca²⁺ + 2Cl⁻ → i = 3

Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻ → i = 5

---

Step 3: Compare i × M

Solution	i × M
I	1 × 0.0976 = 0.0976
II	3 × 0.0684 = 0.205
III	1 × 0.30 = 0.30
IV	5 × 0.08 = 0.40

---

Higher the value of i × M, higher will be the boiling point.

Therefore, increasing order of boiling point:

I < II < III < IV

Related JEE Main Questions