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A mixture of 100m mol of Ca(OH)(2) and ...

A mixture of 100m mol of `Ca(OH)_(2)` and 2g 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 `5.5xx10^(-6)` )

A

13.6 g, 0.14 `"mol L^(-1)`

B

13.6 g, `0.28 "mol"L^(-1)`

C

`1.9 g, 0.28 "mol L"^(-1)`

D

`1.9 g, 0.14 "mol L"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`underset(100 "m mol")(Ca(OH)_(2)) + underset(2000/142)(Na_(2)SO_(4)) to CaSO_(4) + 2OH^(-)`
`= 13.986 = 14` mili mol
mass of `CaSO_(4) = (14 xx 136)/1000 = 1.9 g`
`[OH^(-)] = 28/100 = 0.28 M`
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