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The solubility product of Pbl2 is 8.0×10-6. The solubility of lead idodide in 0.1 molar solution of lead nitrate is x×10−6mol/L. The value of x is ____________( Rounded off to the nearest integer) Given 2=1.41

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answer is 141.

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Detailed Solution

(0.1+s)4 s2=8×10-9=4 s2=8×10-8 S=2×10-4=141.4×10-6=141

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