NCERT Class XI Chemistry Equilibrium Solutions

Question : 72

Total: 73

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6)

Solution:

CaSO4(s) ⇌ Ca(aq)2++SO4(aq)2–
If S is the solubility of CaSO4 in mol‌L–1, then
Ksp = [Ca2+]×[SO42–] = S2
or , S = √Ksp = √.1×10−6 = 3.02 × 10−3molL−1
= 3.02 × 10–3 × 136 gL–1 = 0.411 gL–1
(Molar mass of CaSO4 = 136 gmol–1)
Thus, for dissolving 0.411 g, water required = 1 L
∴ For dissolving 1 g, water required =

0.411

L = 2.43 L

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