7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of `N//15 NaOH` solution for complete neutralisation. Calculate the equivalent mass of the acid.

Explanation

Calculation of the normality of the acid
Volume of acid taken = 20 mL
volume of base required = 25 mL
Normality of base `= N//15`
Applying normality equation : `ubrace(N_(1)xxV_(1))_("Acid") equiv ubrace(N_(2) xx V_(2))_("Base")`
`N_(1) xx (20 mL) = ((1)/(15) N) xx 25 mL`
`N_(1) = ((1)/(15) N) xx ((25 mL))/((20 mL)) = (1)/(12) N`
Step II. Calculation of the equivalent mass of the acid
Amount of acid dissolved per litre of solution = 7.5 g
`:.` Strength of the acid `= 7.5 g L^(-1)`
Equivalent mass of the acid `= ("Strength of the acid")/("Normality of acid") = ((7.5 g L^(-1)))/((1//12 N)) = 90.0`