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

40 g of glucose (Molar mass =180) is mixed with 200 mL of water. The freezing point of solution is __K. (Nearest integer)
[Given : Kf=1.86 K kg mol-1; Density of water =1.00 g cm-3; Freezing point of water =273.15 K

Solution

Molality =40180mol0.2Kg=109 molal

⇒ΔTf=Tf-Tf'=1.86×109

⇒Tf'=273.15-1.86×109

=271.08 K

≈271 K (nearest-integer)

Asked in: JEE Main 2021 (27 Aug Shift 2)

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