Chemistry, asked by anjud5264, 4 months ago

What is the minimum value of water required to dissolved 1 g of calcium sulphate Al 298K ?
(For calcium Sulphate is 9.8 ×10^-9

Answers

Answered by pambade

Answer:

=2.43L.

Explanation:

Let s be the solubility of CaSO4.CaSO4⇌Ca2+SO42−[Ca2+]=[SO42+]=sKsp[Ca2+][SO42−]=s×s=s2=9.1×10−6s=0.003016MThe molar mass of CaSO4 is 40+32+64=136g.The solubility in g/L is 0.003017×136=0.410g/L.Hence, 0.41 g dissolves in 1 L.1 g will dissolve in 0.411=2.43L.

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