JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
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.
- A \(10^7\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^3\)
- B \(10^7\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5\)
- C \(10^3\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5\)
- D \(10^5\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5\)
Answer & Solution
Correct Answer
(B) \(10^7\left(\frac{\mathrm{W}}{\mathrm{M}}\right)^5\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S}=\frac{\mathrm{W} \times 10}{\mathrm{M}}\) \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(\mathrm{~s}) \underset{3 \mathrm{~s}}{\rightleftharpoons} 3 \mathrm{Ca}^{2+}(\text { aq. })+2 \mathrm{PO}_4^{3-}(\text { aq. })\)…
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