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Ksp of AgCl is 10^-10. What is the minimum mass of AgCl(s) required to prepare 100 mL saturated solution of AgCl? [Atomic mass of Ag = 108]

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0.1435 mg

1.435 mg

2.87 mg

7 mg

answer is A.

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Detailed Solution

$K_{s p} = S^{2} \Rightarrow S = {(10^{- 10})}^{1 / 2} = 10^{- 5} M$

$S = 10^{- 5} = \frac{mass of AgCl \times 1000}{143.5 \times 100}$

$Mass of AgCl = 0.143 mg$

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