Consider the following aqueous solutions. I. 2.2 g Glucose in 125...

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

\mathrm{g} \mathrm{mol}^{-1}: \mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Cl}=35.5, \mathrm{Ca}=40, \mathrm{Al}=27

and

\mathrm{S}=32

]

Options

A $\text{ III }<\text{ I }<\text{ II }<$ $\text{ IV }$
B $\mathrm{I}<$ \mathrm{II} $<$ \mathrm{III} $<\mathrm{IV}$
C $\text{ II }<\text{ III }<\text{ I }<$ $\text{ IV }$
D $\text{ II }<\text{ III }<\text{ IV }<$ $\text{ I }$

Correct answer

B. $\mathrm{I}<$ \mathrm{II} $<$ \mathrm{III} $<\mathrm{IV}$

Step-by-step solution

\Delta T_b = i \cdot k_b \cdot m

For dilute solution (

M = m

)

Molarity	$i \times m$
(I) $M_{glucose} = \dfrac{2.2}{180} \times \dfrac{1000}{125} = 0.098$	$0.098 \times 1$
(II) $M_{CaCl_2} = \dfrac{1.9}{111} \times \dfrac{1000}{250} = 0.068$	$0.068 \times 3$
(III) $M_{urea} = \dfrac{9}{60} \times \dfrac{1000}{500} = 0.3$	$0.3 \times 1$
(IV) $M_{Al_2(SO_4)_2} = \dfrac{20.5}{342} \times \dfrac{1000}{750} \approx 0.08$	$0.08 \times 5$

Order of

\Delta 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

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