JEE MAIN - Chemistry (2019 - 10th January Morning Slot - No. 22)
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, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1
, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6
)
13.6g, 0.28 mol L$$-$$1
13.6g, 0.14 mol L$$-$$1
1.9g, 0.28 mol L$$-$$1
1.9g, 0.14 mol L$$-$$1
Explanation
Ca(OH)2 + Na2SO4 $$ \to $$ CaSO4 + 2NaOH
100 m mol 14 m mol $$-$$ $$-$$ $$-$$
$$-$$ $$-$$ $$-$$ 14 m mol 28 m mol
wCasO4 = 14 $$ \times $$ 10$$-$$3 $$ \times $$ 136 = 1.9 gm
[OH$$-$$] = $${{28} \over {100}}$$ = 0.28 M
100 m mol 14 m mol $$-$$ $$-$$ $$-$$
$$-$$ $$-$$ $$-$$ 14 m mol 28 m mol
wCasO4 = 14 $$ \times $$ 10$$-$$3 $$ \times $$ 136 = 1.9 gm
[OH$$-$$] = $${{28} \over {100}}$$ = 0.28 M
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