At 310 ~K, the solubility of CaF₂ in water is 2.34 × 10⁻³ ~g / 100 ~mL. The solubility product of CaF₂ is _ _ _ _ _ _ × 10⁻⁸(~mol / L)³. (Give molar mass: CaF₂=78 ~g ~mol⁻¹)

Equilibrium: JEE Main 2022 Chemistry question

JEE Main 2022 · 27 Jul, Morning shift · Q14

JEE Main 202227 Jul, Morning shiftChemistryMediumEquilibrium123× since 2002

310 \mathrm{~K}

, the solubility of

\mathrm{CaF}_{2}

in water is

2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}

. The solubility product of

\mathrm{CaF}_{2}

is ____________

\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}

. (Give molar mass :

\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}

)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Equilibrium Question with Answer (27 Jul Morning shift, Q14) | JEE Arc