Answer:
Step-by-step explanation:
AgCl ⇄ Ag⁺ + Cl⁻
m m m
If x mole of AgCl be dissolved in one litre .
[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰
m² = 1.6 x 10⁻¹⁰
m = 1.26 x 10⁻⁵ moles
So solubility of AgCl is 1.26 x 10⁻⁵ moles / L
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