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Friday, 9 October

JEE Main 2025 · ChemistryMultiple choiceSingle correctMediumCalculation

JEE Main 8 April 2025, Shift 2, Chemistry Q53

Question 53 of 75 in this shift, Chemistry question 3 of 25, Section A.

HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)} The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given : Kf(H2O)=1.8 K kg mol−1\mathrm{K_f(H_2O)} = 1.8\ \mathrm{K\ kg\ mol^{-1}}, molality ≡ molarity
  1. (1)1.1×10−21.1 \times 10^{-2}
  2. (2)1.38×10−31.38 \times 10^{-3}Official answer
  3. (3)1.89×10−11.89 \times 10^{-1}
  4. (4)1.90×10−31.90 \times 10^{-3}

Official answer

Option 2

NTA final key.

Chapter
Solutions
Topic
Colligative Properties
Idea tested
Calculating Ka from Freezing Point Depression

Same idea in other shifts

Asked 3× in all
  1. 2 Apr 2026, Shift 1 · Q5519.519.5 g of fluoro acetic acid (molar mass =78 g mol−1=78\ \mathrm{g\ mol^{-1}}) is dissolved in 500500 g of water at 298298 K. The depression in the…MediumSingle correct
  2. 29 Jun 2022, Shift 1 · Q831.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid…MediumNumerical value

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 42 of the 174 JEE Main shifts held from 2017 to 2026.