Verified Solution • 2025–26 Session moderate

QJEEM19Ch3ac4: JEE Main 2019 Chemistry - Class 11 JEEM19Ch3ac4 Solution: 100 mL of a water sample contains 0.81 g of calcium bicrabonate …

100 mL of a water sample contains 0.81 g of calcium bicrabonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is : (Molar mass of calcium bicarbonate is 162 g mol–1 and magnesium bicarbonate is 146 g mol–1)

Multiple Choice Options

  • A 1,000 ppm
  • B 100 ppm
  • C 10,000 ppm
  • D 5,000 ppm

Step-by-Step Verified Solution

Here hardness of water is expressed in terms of CaCO3.

∴\therefore Equivalent of CaCO3

= Equivalent of Ca(HCO3)2 + Equivalent of Mg(HCO3)2

⇒\Rightarrow 2 ×\times W100{W \over {100}} = 0.81162×2+0.73146×2{{0.81} \over {162}} \times 2 + {{0.73} \over {146}} \times 2

⇒\Rightarrow W = 1 gm

Volume of water = 100 mL

∴\therefore Mass of water = 100 g

∴\therefore Hardness = 1.0100×106{{1.0} \over {100}} \times {10^6} = 10000 ppm

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring