Concentrated sulphuric acid has density of 1.9 g/mL and 99% `H_(2)SO_(4)` by mass. Calculate the molarity of the acid.

Explanation

Let v mL of 95.6 mass % sulphuric acid of density `1.84 g ml^(-1)` be added to 1000mL of 40 mass % `H_(2)SO_(4)` so that total solution of 50 mass % `H_(2)SO_(4)` becomes (1000 + v)mL.
Volume of solution of 95.6% `H_(2)SO_(4) = (100)/(1.84)`
Amount of `H_(2)SO_(4)` in v mL of solution ` = 95.6 xx (1.84)/(100) xx v`
Volume of solution of 40 % `H_(2)SO_(4) = (100)/(1.31)`
Amount of `H_(2)SO_(4)` in 1000 mL of solution `=(40 xx 1.31)/(100) xx 1000`
Volume of solution of 50% `H_(2)SO_(4) = (100)/(1.40)`
Amount of `H_(2)SO_(4)` in (1000+ v) mL of solution .
`= 50 xx (1.40)/(100) xx (1000+v)`
Thus , `(95.6 xx 1.84 xx v)/(100) + (40 xx 1.31)/(100) xx 1000 = (50 xx 1.40)/(100) (1000+v)`
` 175.90v + 52400 = 70000+70v`
`105.90 v = 17600`
or ` v = 166.2 mL`
vol. of `H_(2)SO_(4)` of density 1.84 contaning 95.6 mass % of `H_(2)SO_(4)` to be added = 166.2 mL