At 310 {\K}, the solubility of {\CaF}_2 in water is 2.34 \times 10^{-3} {\g} \/ 100 {\mL}. The solubility product of {\CaF}_2 is _______ \times 10^{-8}( {\mol} \/ {\L})^3. (Give molar mass : {\CaF}_2=78 {\g} {\mol}^{-1} ) [27-Jul-2022-Shift-1]

Ionic Equilibrium Part 1

Section: Chemistry

Question : 17 of 100

Marks: +1, -0

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

[27-Jul-2022-Shift-1]

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