40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of — JEE Mains Chemistry Past Papers Chemistry Question
Question
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of
Answer: .
💡 Solution & Explanation
Molality = Kg . mol 40 = 10 molal Tf = Tf – Tf’ = 1.86 × 10 Tf’ = 273.15 – 1.86 × 10 = 271.08 K 271 K (nearest-integer)
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