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Q.

The solubility product of PbI2   is   7.2×10−9 Calculate the maximum mass of NaI which may be added in 500 mL of 0.005​ M−Pb(NO3)2solution without any precipitation of PbI2Atomic weight of Na=23, I=127

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a

0.09 g

b

0.9 g

c

9 g

d

90 g

answer is A.

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Detailed Solution

Given,The solubility product of PbI2  ,Ksp=7.2×10−9Volume,v=500 mLmolarity, M = 0.005 M maximum mass of  , m = ? prevent any precipitation of PbI2,Q
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