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`0.9698 g` of an acid are present in `300 mL` of a solution. `10 mL` of this solution requires exactly `20 mL` of `0.05 N KOH` solution. If the `mol.wt.` of acid is `98`, calculate the number of neutralizable protons. |
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Answer» Correct Answer - 3 Meq.of acid in `10 mL` solution `= "Meq.of" KOH` `=20xx0.05` `:.` Meq.of acid in `300 mL` solution `=(20xx0.05xx300)/(10)=30` `:. (0.9698)/(98//n)xx1000=30` ( where `n` is no.of neutralizable proton) `therefore n= 3` |
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