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Q.

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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a

7 mg

b

0.1435 mg

c

1.435 mg

d

2.87 mg

answer is A.

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Detailed Solution

Ksp=S2⇒S=(10−10)1/2=10−5M

S=10−5=mass of AgCl×1000143.5×100

Mass of AgCl=0.143mg

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