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JEE MAIN 2021
24-02-2021 S2
Question
The solubility product of $\mathrm{Pbl}_2$ is $8.0 \times 10^{-9}$. The solubility of lead iodide in 0.1 molar solution of lead nitrate is $x \times 10^{-6} \mathrm{~mol} / \mathrm{L}$. The value of $x$ is $\_\_\_\_$ . (Rounded off to the nearest integer) $[$ Given $: \sqrt{2}=1.41]$
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