JEE Main 2022 · 29 Jul · Shift 1 · Chemistry Q86

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ChemistryInteger+4 / -1Question 86 of 90

If the solubility product of PbS is $8\times10^{-28}$ , then the solubility of PbS in pure water at 298 K is $x\times10^{-16}$ mol L $^{-1}$ . The value of $x$ is _______. (Nearest Integer)

[Given : $\sqrt2$ = 1.41]

Your answer

Answer and worked solution

Official answer: 282

Solubility $s=\sqrt{8\times10^{-28}}=2\sqrt2\times10^{-14}=282\times10^{-16}$ mol/L.

Answer: 282

Topics:Solubility product

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JEE Main 2022 · 29 Jul · Shift 1, Chemistry question 86