JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

100 mL of a water sample contains 0.81 g of calcium bicarbonate 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)

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

Solution

Moles of Ca(HCO₃)₂ = 0.005

Moles of Mg(HCO₃)₂ = 0.005

Hardness in terms of CaCO₃ pm

$= \frac{( 0 .005 + 0 .005 ) \times 100}{100} \times 1 0^{6}$

$= 1 0^{4} ppm$

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