40 g of glucose (Molar mass = 180 ) is mixed with 200 mL of water. The freezing point of solution is _ _K . (Nearest integer) [Given : K f = 1 . 86 K kg mol - 1 ; Density of water = 1 . 00 g cm - 3 ; Freezing point of water = 273 . 15 K

Question

40 g of glucose (Molar mass = 180 ) is mixed with 200 mL of water. The freezing point of solution is _ _K . (Nearest integer) [Given : K f = 1 . 86 K kg mol - 1 ; Density of water = 1 . 00 g cm - 3 ; Freezing point of water = 273 . 15 K,

MARKS App · Accepted Answer

Molality = 40 180 mol 0 . 2 Kg = 10 9 molal ⇒ Δ T f = T f - T f ' = 1 . 86 × 10 9 ⇒ T f ' = 273 . 15 - 1 . 86 × 10 9 = 271 . 08 K ≈ 271 K (nearest-integer)

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