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 sol

Question

Consider the following aqueous solutions. [2026]

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]

Smart Achievers · Accepted Answer

(2)

$Δ T_{b} = i \cdot k_{b} \cdot m$

For dilute solution $(M = m)$

	Molarity	$i \times m$
(I)	$M_{glucose} = \frac{2.2}{180} \times \frac{1000}{125} = 0.098$	$0.098 \times 1$
(II)	$M_{{CaCl}_{2}} = \frac{1.9}{111} \times \frac{1000}{250} = 0.068$	$0.068 \times 3$
(III)	$M_{urea} = \frac{9}{60} \times \frac{1000}{500} = 0.3$	$0.3 \times 1$
(IV)	${M_{{Al}_{2} ({SO}_{4})}}_{3} = \frac{20.5}{342} \times \frac{1000}{750} = 0.08$	$0.08 \times 5$

$Order of Δ T_{b} = {Al}_{2} {({SO}_{4})}_{3} > Urea > {CaCl}_{2} > Glucose$

$So order of BP = {Al}_{2} {({SO}_{4})}_{3} > Urea > {CaCl}_{2} > Glucose$

$So Answer will be I < II < III < IV$

Solution

1 III < I < II < IV

2 I < II < III < IV

3 II < III < I < IV

4 II < III < IV < I

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