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A hard water sample has 131 ppm CaSO_(4). What fraction of the water must be evporated in a container before solid CaSO_(4) begins to deposit. K_(sp) of CaSO_(4) = 9.0 xx 10^(-6).

Answer» <html><body><p></p>Solution :Maximum solubility of `CaSO_(4)` in water <br/> `S = sqrt(K_(sp)) = 3 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-3)mol <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1)` <br/> Let `V` litre of sample is taken, then `CaSO_(4)` present <br/> `= (<a href="https://interviewquestions.tuteehub.com/tag/131-1785688" style="font-weight:bold;" target="_blank" title="Click to know more about 131">131</a> xx V xx 10^(3))/(10^(6)) <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> [`because` ppm = g of `CaSO_(4)` in `10^(6) g` of sample] <br/> `= 131 xx 10^(-3) V g` <br/> `= (131 xx 10^(-3)xxV)/(136)` mole in V L <br/> If water is evaporated on <a href="https://interviewquestions.tuteehub.com/tag/heating-1017297" style="font-weight:bold;" target="_blank" title="Click to know more about HEATING">HEATING</a> so that just precipitation of `CaSO_(4)` occurs. Let `V_(1)L` of water is left, then <br/> `(131 xx 10^(-3)xxV)/(136)` mol is present in `V_(1)L` solution is equal to `3 xx 10^(-3) xx V_(1)` mol <br/> `:. (131 xx 10^(-3) xx V)/(136) = 3 xx 10^(-3) xx V_(1) :. V_(1) = 0.32V` <br/> Thus, volume evaporated =` V - 0.32 V = 0.68V` or `68%` of water should be evaporated.</body></html>


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