JEE Main202628 January 2026Evening ShiftChemistrySolutionsActual
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 ]
Options
- AIII < I < II < IV
- BI < II < III < IV
- CII < III < I < IV
- DII < III < IV < I
Correct answer
B. I < II < III < IV
Step-by-step solution
T_b = i k_b m For dilute solution ( M = m ) Molarity i m (I) M_ glucose = 2.2 180 1000 125 = 0.098 0.098 1 (II) M_ CaCl₂ = 1.9 111 1000 250 = 0.068 0.068 3 (III) M_ urea = 9 60 1000 500 = 0.3 0.3 1 (IV) M_ Al₂(SO₄)₂ = 20.5 342 1000 750 0.08 0.08 5 Order of T_b = Al₂(SO₄)₃ > Urea > CaCl₂ > Glucose So order of BP = Al₂(SO₄)₃ > Urea > CaCl₂ > Glucose So Answer will be I < II < III < IV