Question

Asked in: JEE Main-2019

A 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}

Answers (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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