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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2025 · 3 April · Morning Shift · Chemistry · Solutions

2 moles each of ethylene glycol and glucose are dissolved in 500 g...

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is :
(Given : Ebullioscopic constant of water =0.52 K kg mol−1)\left.=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)

Options

  1. A379.2 K
  2. B377.3 K
  3. C375.3 K
  4. D277.3 K

Correct answer

B. 377.3 K

Step-by-step solution

ΔTb=i1m1kb+i2m2kb=1×20.5×0.52+1×20.5×0.52=4.16(Tb)solution =373.16+4.16=377.3K.\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i}_1 \mathrm{ m}_1 \mathrm{k}_{\mathrm{b}}+\mathrm{i}_2 \mathrm{ m}_2 \mathrm{k}_{\mathrm{b}} \\ & =1 \times \frac{2}{0.5} \times 0.52+\frac{1 \times 2}{0.5} \times 0.52=4.16 \\ & \left(\mathrm{ T}_{\mathrm{b}}\right)_{\text{solution }}=373.16+4.16=377.3 \mathrm{ K}.\end{aligned}
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