`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.

Answer

3

Explanation

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`