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)` )

Options

A. 1.9 g `0.14 mol L^(-1)`
B. 13.6 g 0.14 mol `L^(-1)`
C. 1.9 g 0.28 mol `L^(-1)`
D. 13.6 g, 0.28 mol `L^(-1)`

Answer

C

Explanation

`Ca(OH)_(2) + Na_(2)SO_(4) to CaSO_(4) + 2NaOH`
Moles of `Na_(2)SO_(4) = 2/(143) = 14` m mol
`Na_(2)SO_(4)` is limiting reagent
Weight of `CaSO_(4)` formed `=14 xx 10^(-3) xx 136 = 1.9`g
Moles of NaOH `=28 xx 10^(-3)` mol
Moles of `OH^(-) = 28 xx 10^(-2)` mol
Volume = 100 mL
`[OH^(-)] =(28 xx 10^(-3))/100 xx 1000 = 0.28` M