0.99 g of an acid was dissolved in water and the solution made up to `200 cm^(3)`, `20 cm^(3)` of this solution required 15 `cm^(3)` of 0.105 N sodium hydroxide solution for complete neutralization. Find the equivalent mass of the acid.

Explanation

`underset("Acid")(V_(1) xx N_(1)) = underset("Base")(V_(2) xx N_(2))`
Substituting we have `20 xx N_(1) = 15 xx 0.105` , `therefore N_(1) = (15 xx 0.105)/20 = 0.07875`
`therefore` Normality of acid solution = 0.07875
Mass of the acid in one dm of the solution `=0.99 xx 5 g = 4.95` g
Eq. mass of the acid is given by `E= W/N = 4.95/0.07875 = 62.85`