83 g of ethylene glycol dis | JEE Mains Chemistry Past Papers — Chem-Mantra | Chem-Mantra

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Question

83 g of ethylene glycol dis

Answer: .

💡 Solution & Explanation

Tƒ = iKƒ × m = 62 83 . 1     Tƒ = 3.98 So, freezing point of solution = 273 – 3.98 = 269.02 = 269 K | JEE MAIN-2021 | DATE : 26-08-2021 (SHIFT-2) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 10

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