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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Hydrogen

JEE Main 2019 · 12 January · Morning Shift · Chemistry · Hydrogen

What is the hardness of a water sample ( in terms of equivalents of...

What is the hardness of a water sample ((in terms of equivalents of CaCO3){CaCO}_{3} ) containing 10−3MCaSO4{10}^{- 3}M{CaSO}_{4}?

((Molar mass of CaSO4=136gmol−1){CaSO}_{4} = 136g{mol}^{- 1} )

Options

  1. A9090 ppm
  2. B5050 ppm
  3. C1010 ppm
  4. D100100 ppm

Correct answer

D. 100100 ppm

Step-by-step solution

As 1L1L solution have 10−3molCaSO4{10}^{- 3}mol{CaSO}_{4}

Eq.ofCaSO4=eq.ofCaCO3Eq . of{CaSO}_{4} = eq . of{CaCO}_{3}

In 1L1L solution

nCaSO4×v.f=nCaCO3×v.f.{n}_{{CaSO}_{4}} \times v . f = {n}_{{CaCO}_{3}} \times v . f .

10−3×2=nCaCO3×2{10}^{- 3} \times 2 = {n}_{{CaCO}_{3}} \times 2

nCaCO3=10−3molin1L{n}_{{CaCO}_{3}} = {10}^{- 3}molin1L

∴WCaCO3=100×10−3g∴ {W}_{{CaCO}_{3}} = 100 \times {10}^{- 3}g in 1L1L solution

∴∴ hardness in terms of CaCO3{CaCO}_{3}

=WCaCO2Wtotal×106=100×10−3g1000g×106=100ppm= \dfrac{{W}_{{CaCO}_{2}}}{{W}_{total}} \times {10}^{6} = \dfrac{100 \times {10}^{- 3}g}{1000g} \times {10}^{6} = 100ppm
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