JEE Main 2022 · 26 June · morning · Single correct · +4 / −1

A 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:
(Atomic mass : Cl = 35.5 amu, H = 1 amu)

  1. (a) 10.2 M
  2. (b) 12.5 M
  3. (c) 14.0 M
  4. (d) 18.2 M
Answer and solution

Answer (c) 14.0 M

Solution

Chemistry ma’am · 3 steps

Take exactly 1 L of the acid; the density gives its mass, and the 35% gives the HCl inside it.

  1. Mass of 1 L. Molarity is moles of solute per litre of solution, so work with 1000 mL. Its mass is 1000 × 1.46 = 1460 g.
  2. Mass of HCl. 35% of that mass is HCl: 0.35 × 1460 = 511 g. The molar mass of HCl is 1 + 35.5 = 36.5 g mol⁻¹.

    M = (511 g)/(36.5 g mol⁻¹ × 1 L) = 14.0 mol L⁻¹

  3. So. The molarity is 14.0 M, which is option (c). The shortcut M = (10 × d × w%)/(molar mass) = (10 × 1.46 × 35)/36.5 gives the same answer.

The idea to keep. Mass percent and density together give molarity. Fix the volume at 1 L, use density to get its mass, then take the percent of that mass as solute.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Mole Concept & Stoichiometry

Asked about 210 times in JEE Main since 2020. This question is from 2022.