JEE Main 2021 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 27 August 2021, Shift 2, Chemistry Q56

Question 56 of 90 in this shift, Chemistry question 26 of 30, Section B.

40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given :

\mathrm{K_f}

= 1.86 K kg

\mathrm{mol^{-1}}

; Density of water = 1.00 g

\mathrm{cm^{-3}}

; Freezing point of water = 273.15 K]

Official answer

271

NTA final key.

Chapter: Solutions
Topic: Colligative Properties
Idea tested: Freezing Point Depression

Same idea in other shifts

Asked 16× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.