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 $x \times 10^{-8} \, (mol / L)^{3}$. Find the value of $x$. (Given molar mass: $CaF_{2} = 78 \, g \, mol^{-1}$)

  • A
    $0.0108$
  • B
    $0.108$
  • C
    $1.08$
  • D
    $10.8$

Explore More

Similar Questions

If the solubility product of $AgCl$ is $10^{-10}$,what is the solubility of $AgCl$ in $0.001 \ M \ NaCl$?

The molar solubility of $PbI_2$ in $0.2 \ M \ Pb(NO_3)_2$ solution in terms of $K_{sp}$ (solubility product) is

The solubility product of $AgCl$ is $1.8 \times 10^{-10}$. Mixing equal volumes of which of the following solutions will result in the precipitation of $AgCl$?

Difficult
View Solution

The solubility product of $AgI$ at $25\,\text{°C}$ is $1.0 \times 10^{-16}\,\text{mol}^2\,\text{L}^{-2}$. The solubility of $AgI$ in $10^{-4}\,\text{N}$ solution of $KI$ at $25\,\text{°C}$ is approximately (in $\text{mol L}^{-1}$)

In an aqueous solution,$SCN^-$,$Br^-$,$I^-$,and $Cl^-$ are present. Which will get precipitated first when $AgNO_3$ is added to the solution? Given that:
$K_{sp}(AgCl) = 1.2 \times 10^{-10}$,
$K_{sp}(AgI) = 1.7 \times 10^{-16}$,
$K_{sp}(AgSCN) = 7.1 \times 10^{-7}$,
$K_{sp}(AgBr) = 3.5 \times 10^{-13}$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo