2.029 * 10-3mol dm-3
1.414 * 10-3mol dm-3
2.029 * 10-5mol dm-3
1.414 * 10-5mol dm-3
Correct answer is D
In order to calculate this, it suffices to calculate the solubility of Ag, so as to know the amount lost.
\(K_{sp} = [Ag^{+}][Cl^{-}]\)
Let the solubilty of Ag and Cl = d
\(2 \times 10^{-10} = d \times d= d^{2}\)
\(d = \sqrt{2\times10^{-10}}\)
= \(1.414 \times 10^{-5}moldm^{-3}\)
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