Solubility of calcium phosphate (molecular mass, M ) in water is W...

Given the solubility of ${Ca}_{3}{\left({PO}_{4}\right)}_{2}$ is $Wg / 100mL$. The solubility in $mol / L$ is $S = \dfrac{W}{M} \times \dfrac{1000}{100}mol / L$.

The equilibrium reaction of calcium phosphate is 

$\begin{cases} & {Ca}_{3}{\left({PO}_{4}\right)}_{2} & \overset{}{⇌} & 3{Ca}^{2 +} & + & 2{PO}_{4}^{3 -} \\ Atequilibrium & & & 3S & & 2S\end{cases}$

Now, ${K}_{sp} = {\left[{Ca}^{2 +}\right]}^{3}{\left[{PO}_{4}^{3 -}\right]}^{2}$

$⇒ {K}_{sp} = {\left(3S\right)}^{3}{\left(2S\right)}^{2} = 108 \times {S}^{5}$

$⇒ {K}_{sp} = 108 \times {\left(\dfrac{W}{M} \times 10\right)}^{5} ⇒ {K}_{sp} = 108 \times {10}^{5} \times {\left(\dfrac{W}{M}\right)}^{5}$

$= 1 . 08 \times {10}^{7}{\left(\dfrac{W}{M}\right)}^{5}$