JEE Mains Chemistry Past PapershardNUMERICAL

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