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A sample of hard water contains 100 ppm of CaSO_(4).What minimum fraction of water shouldbe evaporated off so that solid CaSO_(4) begins to separate out ? K_(sp)for CaSO_(4)is 9.0xx10^(-6).

Answer» <html><body><p></p>Solution :Maximum solubility of `CaSO_(4)` in water can be calculated from its `K_(sp)` value as <a href="https://interviewquestions.tuteehub.com/tag/follows-994526" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWS">FOLLOWS</a>: <br/> `S=sqrt(K_(sp))=sqrt(9.0xx10^(-6))=3.0=10^s(-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/> Suppose the volume of water taken = V litre <br/> As `CaSO_(4)` present is 100 ppm, i.e., 100 g per `10^(6) g ` of water, therefore, `CaSO_(4)` present in V litres `(V xx 10^(3)g)` of water <br/> `=(100)/(10^(6))xxV xx 10^(3) g = 0.1 V g = (0.1 V)/(136)` moles(Molar mass of `CaSO_(4) = 136` g `"mol"^(-1)`) <br/> After evaporation, suppose volume of water left = V' litre <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, V' litre of water will now contain `= (0.1 V) /(136) ` moles of `CaSO_(4)`. <br/> This should be <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> to the maximum solubility in moles `L^(-1)`. <br/> `:. (0. V)/(136) xx (1)/(V')= 3.0 xx 10^(-3) or V' = (0.1 V)/(136 xx 3 xx 10^(-3))=0.245 V ` <br/> `:. ` Volume of water <a href="https://interviewquestions.tuteehub.com/tag/evaporated-7679270" style="font-weight:bold;" target="_blank" title="Click to know more about EVAPORATED">EVAPORATED</a>`= V - V' = V - 0.245 V = 0.755 V` <br/> i.e., 75.5 % of water should be evaporated off .</body></html>


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