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Question: The solubility product of AgCl is \(1.5625 \times 10 ^ { - 10 }\) at \(25 ^ { \circ } \mathrm { C ...

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

A

143.5143.5

B

108

C

1.57×10−81.57 \times 10 ^ { - 8 }

D

1.79×10−31.79 \times 10 ^ { - 3 }

Answer

1.79×10−31.79 \times 10 ^ { - 3 }

Explanation

Solution

: AgCl Ag+s+Cl−s\underset { \mathrm { s } } { \mathrm { Ag } ^ { + } } + \underset { \mathrm { s } } { \mathrm { Cl } ^ { - } }

s2=1.5625×10−10\mathrm { s } ^ { 2 } = 1.5625 \times 10 ^ { - 10 }

s=1.25×10−5 mol L−1\mathrm { s } = 1.25 \times 10 ^ { - 5 } \mathrm {~mol} \mathrm {~L} ^ { - 1 }

Solubility in g L−1=\mathrm { L } ^ { - 1 } = Molar mass ×s

=143.5×1.25×10−5=1.79×10−3gL−1= 143.5 \times 1.25 \times 10 ^ { - 5 } = 1.79 \times 10 ^ { - 3 } \mathrm { gL } ^ { - 1 }