JEE Main 2021 — Solutions Question with Solution
From: JEE Main 2021 (Online) 27th August Evening Shift
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 : Kf = 1.86 K kg mol1; Density of water = 1.00 g cm3; Freezing point of water = 273.15 K]
Enter your answer
Show full solutionCorrect answer: 271.08
Correct answer
271.08
Step-by-step explanation
Molality = molal
= 271.08 K
271 K (nearest - integer)
= 271.08 K
271 K (nearest - integer)
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This is a previous-year question from JEE Main 2021, covering the Solutions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.