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A mixture of formic acid and oxalic acid is heated with conc. H_(2)SO_(4) The case produced is collected and on its treatment with KOH solution, the volume of the gas decreased by 1/6 th. Calculated the molar ratio of the two acids in the original mixture. |
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Answer» Solution :SUPPOSE no. of moles of formic acid is x and that of oxalic acid is y. Then `underset(x)("mole")overset(Conc.)underset(H_(2)SO_(4))rarrH_(2)O_(4)(l)+underset("x mole")(CO(g))` `underset("y mole"){:(" "COOH),("|"),(" "COOH):}rarr underset("y mole")(CO(g))+underset("y mole")(CO_(2)(g))+H_(2)O(l)` Total no. of moles of gaseous product `=x+2y` As only `CO_(2)` is ABSORBED by KOH, therefore, fraction of `CO_(2)=(y)/(x+2y)=(1)/(6)" (GIVEN)"` `"or"6y=x+2y or 4y=x or (x)/(y)=(4)/(1)` Hence, molar ratio of `HCOOH` to `(COOH)_(2)` is `4:1` |
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