JEE Main 2021 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 24 February 2021, Shift 1, Chemistry Q51

Question 51 of 90 in this shift, Chemistry question 21 of 30, Section B.

4.5 g of compound A (MW=90) was used to make 250 mL of its aqueous solution. The molarity of the solution in M is

x \times 10^{-1}

. The value of

x

is __________. (Rounded off to the nearest integer)

Official answer

NTA final key.

Chapter: Some Basic Concepts of Chemistry
Topic: Mole Concept
Idea tested: Molarity Calculation

Same idea in other shifts

Asked 16× in all

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26 Jul 2022, Shift 1 · Q85When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The…MediumNumerical value
26 Jun 2022, Shift 1 · Q61A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is :…EasySingle correct
4 Aug 2021, Shift 1 · Q51A solution of $\mathrm{H_2SO_4}$ has 30% $\mathrm{H_2SO_4}$ by mass and a density of 1.25 g mL $^{-1}$ . The molarity of the solution is…EasyNumerical value
26 Aug 2021, Shift 2 · Q51100 mL of $\mathrm{Na_3PO_4}$ solution contains 3.45 g of sodium. The molarity of the solution is __________ $\times 10^{-2}$ mol…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 108 of the 174 JEE Main shifts held from 2017 to 2026.