A piece of `Al` wieghing `2.7 g` is titrated with `75.0 mL` of `H_(2) SO_(4)` (specific gravity `1.8 mL^(-1)` and 24.7% `H_(2) SO_(4)` by weight). After the metal is completely dissolved, the solution is diluted to `400 mL`. Calculate the molarity of free `H_(2) SO_(4)` solution.

Explanation

Mass of `H_(2)SO_(4)=(24.7)/(100)xx75xx1.18`
=21.8595 g
Reaction : `underset(2xx27)(2Al)+underset(3xx98)(3H_(2)SO_(4)) to Al_(2)(SO_(4))_(3)+3h_(2)`
`H_(2)SO_(4)` required for dissolving 2.7 g Al
`=(3xx98)/(2xx27)xx2.7=14.7 g`
`H_(2)SO_(4)` left unreacted =(21.895-14.7)g=7.1595 g
7.1595 g `H_(2)SO_(4)` is present in 400 mL
Amt. of `H_(2)SO_(4)` present in one litre`=(7.1595)/(400)xx1000 g`
=17.898 g
No. of g moles of `H_(2)SO_(4)=(17.898)/(98)=0.1826`
Hence, molartiy of `H_(2)SO_(4)=0.1826 M`