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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2019 · 10 January · Morning Shift · Chemistry · Ionic Equilibrium

A mixture of 100mmol of Ca ( OH ) _ 2 and 2g of sodium sulphate was...

A mixture of 100mmol100mmol of Ca(OH)2Ca{( OH )}_{2} and 2g2g of sodium sulphate was dissolved in water and the volume was made up to 100mL.100mL . What is the mass of calcium sulphate formed and the concentration of OH−{OH}^{-} in resulting solution, respectively? (Molar mass of Ca(OH)2,Na2SO4Ca{\left(OH\right)}_{2}, {Na}_{2}{SO}_{4} and CaSO4{CaSO}_{4} are 74,14374, 143 and 136gmol−1,136g{mol}^{- 1}, respectively; Ksp{K}_{sp} of Ca(OH)2is5.5×10−6Ca{\left(OH\right)}_{2}is5 .5 \times {10}^{- 6})A mixture of 100mmol of Ca ( OH ) _ 2 and 2g of sodium sulphate was dissolved in water and the volume was made up to

Options

  1. A1.9g,0.14molL−11 .9 g, 0 .14 mol{L}^{- 1}
  2. B13.6g,0.28molL−113 .6 g, 0 .28 mol{L}^{- 1}
  3. C1.9g,0.28molL−11 .9 g, 0 .28 mol{L}^{- 1}
  4. D13.6g,0.14molL−113 .6 g, 0 .14 mol{L}^{- 1}

Correct answer

C. 1.9g,0.28molL−11 .9 g, 0 .28 mol{L}^{- 1}

Step-by-step solution

A mixture of 100mmol of Ca ( OH ) _ 2 and 2g of sodium sulphate was dissolved in water and the volume was made up to

Mass of CaSO4=14×1361000=1.9g{CaSO}_{4} = \dfrac{14 \times 136}{1000} = 1 .9 g

CaSO4=14×1361000=1.9g{CaSO}_{4} = \dfrac{14 \times 136}{1000} = 1 .9 g

\left[{OH}^{-}\right] = \dfrac{28}{100} = 0 .28 M

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