Chemistry · 2021
JEE Main · 24 February 2021, Shift 2 · Q55
The solubility product of PbI_2 is 8.0 × 10^-9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^-6 mol / L. The value…
The solubility product of $\displaystyle \mathrm{PbI}_2$ is $\displaystyle 8.0 \times 10^{-9}$. The solubility of lead iodide in $\displaystyle 0.1$ molar solution of lead nitrate is $\displaystyle x \times 10^{-6} \mathrm{~mol} / \mathrm{L}$. The value of $\displaystyle x$ is \_\_\_\_. (Rounded off to the nearest integer)
[Given $\displaystyle \sqrt{2}=1.41$ ]
Official answer
141
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