JEE Main 2025 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 8 April 2025, Shift 2, Chemistry Q71

Question 71 of 75 in this shift, Chemistry question 21 of 25, Section B.

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is __________ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol

^{-1}

)

Official answer

NTA final key.

Chapter: Some Basic Concepts of Chemistry
Topic: Stoichiometry
Idea tested: Gravimetric Precipitation Analysis

Same idea in other shifts

Asked 4× in all

4 Apr 2026, Shift 2 · Q75In sulphur estimation, $2.0\times10^{-3}$ mol of an organic compound (X) (molar mass 76 $\mathrm{g\ mol^{-1}}$ ) gave 0.4813 g of barium…EasyNumerical value
5 Apr 2026, Shift 1 · Q72In an estimation of sulphur by Carius method 0.2 g of the substance gave 0.6 g of $\mathrm{BaSO_4}$ . The percentage of sulphur in the…EasyNumerical value
13 Apr 2023, Shift 2 · Q900.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is…EasyNumerical 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.