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Q.

The solubility product of AgCl is 10−10. The minimum volume (in L) of water required to dissolve 1. 722 mg of AgCl (molecular weight of AgCl = 143.5) :

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a

20 lt

b

2.2 it

c

1.2 lt

d

10 lt

answer is C.

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Detailed Solution

s=ksp=10−5 M

⇒10−5 moles =1.722×10−3143.5×V(L) ⇒V=1.2L

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The solubility product of AgCl is 10−10. The minimum volume (in L) of water required to dissolve 1. 722 mg of AgCl (molecular weight of AgCl = 143.5) :