# QuestionAsked in: JEE Main-2019A mixture of 100 m mol of $Ca(OH)_{2}$ and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL .The mass of calcium sulphate formed and the concentration of $OH^{-}$ in resulting solution , respectively , are  : (Molar mass of  $Ca(OH)_{2}$  ,$Na_{2}SO_{4} \: and \: \: CaSO_{4}$ are 74, 143 and 136 g $mol^{-1}$ , respectively ; $K_{sp} \: \: of\: \: Ca(OH)_{2}\: \: is\: \: 5.5\times 10^{-6}$)A.$1.9\: g,\: \: 0.28\: mol\: L^{-1}$B.$13.6g,\: \: 0.28\: molL^{-1}$C.$1.9g,\: \: 0.14molL^{-1}$D.$13.6g,\: \: 0.14molL^{-1}$

Na2SO4 + Ca(OH)2 → CaSO4 + 2 NaOH

mmol of Na2SO4 = 2*1000/143

=13.98 m Mol

mmol of CaSO4 formed = 13.98 m Mol

Mass of CaSO4 formed = 13.98 × 10^-3 × 136 = 1.90 g

mmol of NaOH = 28 mmol

Ca(OH)2 → Ca2+ + 2 OH-

[OH–] = 28 mmol / 100ml=0.028 mol/0.1 L =0.28mol L-1

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