The solubility product of {Pbl}_{2} is 8.0 \times 10^{-9}. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x \times 10^{-6} {mol} \/ {L}. The value of x is ....... (Rounded off to the nearest integer). [Given, : \sqrt{2}=1.41 ] [24 Feb 2021 Shift 2]

Ionic Equilibrium Part 1

Section: Chemistry

Question : 43 of 100

Marks: +1, -0

The solubility product of Pbl2 is 8.0×10−9. The solubility of lead iodide 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 ]

[24 Feb 2021 Shift 2]

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