InterviewSolution
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InterviewSolution
| 1. |
In thefraph given below, identify the states of solution atthevarious points A, B, C and E.If thesolution is cooled from point 'A' at which temperature, precipitation normally starts ? Also, find out theamount of solute precipitated at 40^(@)C at A and the amount of solute in thesolution at point 'E'. What would be themaximum amount of solute that can be precipitated in the process? |
| Answer» Solution :Thecurve indicates that thesolubility of thesolute at `100^(@)C` is 250 g. However, the given solution contains only 150 g of solute. THESOLUTION atpoint 'A' is an unsaturated solution. Therefore, no precipitation takes place up topoint 'B' at `60^(@) C`, where thesolution becomes saturated, withjust 150 gas SOLUBILITY. Point 'B'indicates saturated solution andhence, precipitations normally starts at `60^(@)C`. With the fall intemperature, solubility decreases andthe extra amount should get precipitated: Point 'C' at `40^(@)C` indicates that the solution still contained 150 g of solute, whichis a supersaturated solution. THATMEANS, it is possible to cool a solution from `60^(@)C" to" 40^(@)C` without any crystallisation by creating conditions such as absence ofdust particles. At `40^(@)C`, when crystallisation starts, thesolution ULTIMATELY becomes saturated , and this is represented by point D.At pointD (`40^(@)C`), thesolubility is 100 g . That means, 50 g of solute would havebeen precipitated. At 'E' also thesolution remains saturated by precipitating 25 g more ofsolute leaving behing 75 g of solute in the solution. At `0^(@)C`, when water starts to solidify, thesolution still contains 5 g ofsolute without precipitation. That means, 125 g of solute can be precipitated. | |