question_answer 1)

What is the minimum volume of Water required to dissolve 1g calcium sulphate at\[298K?\]For \[CaS{{O}_{4}}\],\[{{K}_{sp}}\] is\[9.1\times {{10}^{-6}}\].

Answer:

Let solubility of \[CaS{{O}_{4}}\] at 298 K is s\[mol\,{{L}^{-1}}\]. Then \[{{K}_{sp}}={{s}^{2}}\] \[s=\sqrt{9.1\times {{10}^{-6}}}=3.016\times {{10}^{-3}}mol/litre\] Solubility in \[g/L\,=3.016\times {{10}^{-3}}\times 136=0.41\,g/L\] \[\therefore \] Minimum volume of water required to dissolve \[1\,g\,CaS{{O}_{4}}=\frac{1000}{0.41}=2439\,mL\] \[=2.439\,L\]