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

Question

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]

(A) I < II < III < IV
(B) III < I < II < IV
(C) II < III < I < IV
(D) II < III < IV < I

Prashant Kotian · Accepted Answer

Correct answer: A. ution. 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] (1) I < II < III < IV (2) III < I < II < IV (3) II < III < I < IV (4) II < III < IV < I Ans. (1) Sol. Tb = i.kb.m For dilute solution (M = m) Molarity i × m (I) Mglucose = 2.2 180  = 0.098 0.098 × 1 (II) CaCl 1.9 M 0.068 250    0.068 × 3 (III) Murea = 9 0.3 500   0.3 × 1 (IV) 4 2 Al (SO ) 20.5 M 0.08 750   0.08 × 5 Order of Tb = Al2(SO4)3 > Urea > CaCl2 > Glucose So order of BP = Al2(SO4)3 > Urea > CaCl2 > Glucose So Answer will be I < II < III < IV

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