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Solubility of calcrum phosphate (molecular mass, M) in water is $\mathrm{W}_{\mathrm{g}}$ per $100 \mathrm{~mL}$ at $25^{\circ} \mathrm{C}$. Its solubility product at $25^{\circ} \mathrm{C}$ will beapproximately.
$10^7\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^3$
$10^7\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5$
$10^3\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5$
$10^5\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5$
Solution
$\mathrm{S}=\frac{\mathrm{W} \times 10}{\mathrm{M}}$
$\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(\mathrm{~s}) \underset{3 \mathrm{~s}}{\rightleftharpoons} 3 \mathrm{Ca}^{2+}(\text { aq. })+2 \mathrm{PO}_4^{3-}(\text { aq. })$
$\mathrm{S}=\frac{\mathrm{W} \times 1000}{\mathrm{M} \times 100}=\frac{\mathrm{W} \times 10}{\mathrm{M}}$
$\mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~s})^3(2 \mathrm{~s})^2$
$=108 \mathrm{~s}^5$
$=108 \times 10^5 \times\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^{\mathrm{s}}$
$=1.08 \times 10^7\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5$
