JEE Main 2022 · 27 Jul · Shift 1 · Chemistry Q86

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ChemistryInteger+4 / -1Question 86 of 90

At 310 K, the solubility of $\mathrm{CaF_2}$ in water is $2.34\times10^{-3}$ g/100 mL. The solubility product of $\mathrm{CaF_2}$ is ______ $\times10^{-8}$ (mol/L) $^3$ . (Give molar mass : $\mathrm{CaF_2}$ = 78 g mol $^{-1}$ )

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Answer and worked solution

Official answer: 0

Solubility $\frac{2.34\times10^{-2}\text{ g/L}}{78}=3\times10^{-4}$ M.

$K_{sp}=4s^3$ $=1.08\times10^{-10}=0.0108\times10^{-8}$ , which rounds to 0.

Answer: 0

Topics:Solubility product

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JEE Main 2022 · 27 Jul · Shift 1, Chemistry question 86