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A solution containing 4.2 g of KOH and `Ca(OH)_(2)` is neutralised by an acid. If it consumes 0.1 g equivalents of the acid, calculate the composition of the sample. |
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Answer» Let mass of KOH be present in mixture=a g and Mass of `Ca(OH)_(2)=(4.2-a)g` Eq. mass of KOH=56, Eq. mass of `Ca(OH)_(2)=(74)/(2)=37` g equivalent of KOH + g equivalent of `Ca(OH)_(2)` = g equivalent of the acid `(a)/(56)+((4.2-a))/(37)=0.1` or `37a-56a=0.1xx56xx37-4.2xx56` or 19a=28 `a=(28)/(19)=1.47` Mass of KOH in the sample =1.47 g Percentage of KOH = 35 and Percentage of `Ca(OH)_(2)=100-35=65` |
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