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Engineering
3 months ago

A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate

Screenshot_811.png A mixture of 100 mmol 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 the solution, respectively, are (Molar mass of Ca(OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6)
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A Ankit
Answered 3 months ago

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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