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Question

Chemistry Question on Equilibrium

KspK_{sp} of CaSO45H2OCaSO_4\cdot5H_2O is 9 x 10610^{-6}, find the volume for 1 g of CaSO4CaSO_4 (M.wt. = 136).

A

2.45 litre

B

5.1 litre

C

4.52 litre

D

3.2 litre

Answer

2.45 litre

Explanation

Solution

CaSO4(s)<=>Ca2+S(aq)+SO42S(aq)CaSO_{4} (s) {<=>} \underset{S}{Ca^{2+}}(aq) +\underset{S}{SO^{2-}_{4}}(aq) Ksp=S2=9×106K_{s p}=S^{2}=9 \times 10^{-6} S=3×103molL1S=3 \times 10^{-3} \,mol \,L ^{-1} Solubility in glitre1g \, litre ^{-1} molecular mass ×S\times S =136×3×103=408×103gL1=136 \times 3 \times 10^{-3}=408 \times 10^{-3} g \,L ^{-1} 408×103g408 \times 10^{-3} \,g of CaSO4CaSO _{4} present in 1 litre 1g1 \,g of CaSO4CaSO _{4} present is 1408×103=2.45\frac{1}{408 \times 10^{-3}}=2.45 litre