JEE Main 2019 — Hydrogen Question with Solution
From: JEE Main 2019 (Online) 8th April Morning Slot
Question
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)
Choose an option
Show full solutionCorrect option: C
Correct answer
C10,000 ppm
Step-by-step explanation
Here hardness of water is expressed in terms of CaCO3.
Equivalent of CaCO3
= Equivalent of Ca(HCO3)2 + Equivalent of Mg(HCO3)2
2 =
W = 1 gm
Volume of water = 100 mL
Mass of water = 100 g
Hardness = = 10000 ppm
Equivalent of CaCO3
= Equivalent of Ca(HCO3)2 + Equivalent of Mg(HCO3)2
2 =
W = 1 gm
Volume of water = 100 mL
Mass of water = 100 g
Hardness = = 10000 ppm
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This is a previous-year question from JEE Main 2019, covering the Hydrogen chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.