Calculate the molality and molarity of a solution made by mixing equal volumes of `30%` by weight of `H_(2)SO_(4)` (density`=1.20g//ml`) and `70%` by weight of `H_(2)SO_(4)` (density`=1.60g//mL`)

Answer

Molality `=11.44 m`, Molarity `=7.55M`

Explanation

Suppose that the solution contains `100 ml` of each variety of `H_(2)SO_(4)`. Total solution is, `200 ml` or `0.2` litre `wt.` of `100 ml` of `H_(2)SO_(4)` solution `(30%)=1060xx100=160` gram
`wt.` of `H_(2)SO_(4)(30%)=120xx(30)/(100)=36` gram
`wt.` of `H_(2)SO_(4)(70%)=160xx(70)/(100)=112` gram
Total `wt.` of `H_(2)SO_(4)` (salute) `=36+112=148` gram
`therefore wt.` of `H_(2)O`(solvent) `=wt.` of solution`-wt.` of solute
`=(120+160)-148`
`=280-148=132` gram
Moles of `H_(2)SO_(4)=(148)/(98)=1.51` (mol `wt.` of `H_(2)SO_(4)=98`)
Molality `=(1.51)/(132)xx1000=11.44m`
Molarity `=(1.51)/(0.2)=7.55 M`