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A mixture of 200 mmol of Ca(OH)2 and 3...

A mixture of 200 mmol of `Ca(OH)_2 ` and 3g of sodium sulphate was dissolved in water and the volume was made up to 200 mL. The mass of calcium sulphate formed and the concentration of ` OH^(-)` in resulting solution, respectively, are: (Molar mass of `Ca(OH)_2 , Na_2SO_4 " and " CaSO_4` are 74, 143 and `136 g mol^(-1)` respectively, `K_(sp)` of `Ca (OH)_2` is ` 5 xx 10^(-6) ` )

A

`2.9 gm 0.21 mol L^(-1)`

B

`13.6 gm , 0.21 mol L^(-1)`

C

`2.9 gm , 0.11 mol L^(-1)`

D

`13.6 gm , 0.11 molL^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(179m ."mole")underset(200m."mole")(Ca(OH)_2) + underset(-)underset(21m."mole")(Na_2SO_4) to CaSO_4 + underset(21m. "Mole" )underset(-)(2NaOH) underset(42m."mole")underset(-)(" ")`
` W_(CaSO_4) = 21 xx 10^(-3) xx 136 = 2.9 gm , [OH^(-) ] (n_(OH^(-)))/(V_("solution")) = 42/200 = 0.21 M`
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