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JEE Main 202127 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
27 August
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2021 · 27 August · Evening Shift · Chemistry · Solutions

40g of glucose (Molar mass = 180 ) is mixed with 200mL of water....

40g40g of glucose (Molar mass =180)= 180 ) is mixed with 200mL200mL of water. The freezing point of solution is __K. (Nearest integer)
[Given : Kf=1.86Kkgmol−1;{K}_{f} = 1 . 86Kkg{mol}^{- 1}; Density of water =1.00gcm−3;= 1 . 00g{cm}^{- 3}; Freezing point of water \left= 273 . 15K\right]

Correct answer

0

Step-by-step solution

Molality = \dfrac{\left(\dfrac{40}{180}\right)\text{ mol }}{0 . 2Kg} = (109)\left(\dfrac{10}{9}\right) molal

⇒ΔTf=Tf−Tf′=1.86×109⇒ \Delta{T}_{f} = {T}_{f} - {T}_{f}^{'} = 1 . 86 \times \dfrac{10}{9}

⇒Tf′=273.15−1.86×109⇒ {T}_{f}^{'} = 273 . 15 - 1 . 86 \times \dfrac{10}{9}

=271.08K= 271 . 08K

≈271K\approx 271K (nearest-integer)

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