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Question

Chemistry Question on Solutions

The solubility product of AgClAgCl is 1.5625×10−101.5625 \times 10^{-10} at 25 ∘C25 \,^{\circ}C. Its solubility in grams per litre will be

A

143.5143.5

B

108108

C

1.57×10−81.57\times10^{-8}

D

1.79×10−31.79\times10^{-3}

Answer

1.79×10−31.79\times10^{-3}

Explanation

Solution

AgCl<=> {AgCl<=>} Ag+s+\underset{\text{s}}{ {Ag+}}+ Cl−s\underset{\text{s}}{ {Cl-}} s2=1.5625×10−10s^{2}=1.5625\times10^{-10} s=1.25×10−5 mol L−1s=1.25\times10^{-5}\,mol\,L^{-1} Solubility in g L−1=g\,L^{-1}= Molar mass ×s\times s =143.5×1.25×10−5=143.5\times1.25\times10^{-5} =1.79×10−3 g L−1=1.79\times10^{-3}\,g\,L^{-1}