Q3 / 75 · JEE Main 2026 (28 January Shift 2) · Chemistry

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 $\mathrm{S}=32$]

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

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

Hence, the correct answer is option: 'B'.

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