Question

At 291 K, the conductivity of saturated solution of $Ca{F}_2$ is $3.86\times {10}^{-5}Sc{m}^{-1}$ and that of water used for solution is $0.15\times {10}^{-5}Sc{m}^{-1}$. The ionic conductance of $C{a}^{2+}$ and $F^{-}$ at infinite dilution are 51.0 and 47.0 $Sc{m}^{2}mo{l}^{-1}$ respectively. The solubility of $Ca{F}_2$ in solution is $X\times {10}^{-2}glitr{e}^{-1}$. Value of $X$ is:

Answer 1

The specific conductivity of $Ca{F}_2$ is the difference in the specific conductivity of saturated solution
and the specific conductivity of water used for the solution.
The specific conductivity $\kappa =3.86\times {10}^{-5}-0.15\times {10}^{-5}=3.71\times {10}^{-5}S/cm$
The molar conductivity at infinite dilution is $\Lambda =51.0+2\times 47.0=145Sc{m}^2/mol$
The molar solubility is $S=\frac{1000\kappa }{\Lambda }=\frac{1000\times 3.71\times {10}^{-5}}{145}=25.6\times {10}^{-4}mol/L$
The molar mass of $Ca{F}_2$ is 78 g/mol.
The solubility in g/L is $78\times 2.56\times {10}^{-4}=0.02g/L$
But it is equal to $x\times {10}^{-2}g/L$
Hence, $x=2$