A sample of `H_(2)SO_(4)` `("density" 1.787 g mL^(-1))` is labelleed as `80%` by weight. What is molarity of acid? What volume of acid has to be used to make `1"litre of" 0.2MH_(2)SO_(4)`?

Explanation

`H_(2)SO_(4)` is `80%` by weight
`:. Wt. of H_(2)SO_(4)= 80 g`
W.t. of solution `= 100 g`
`:.` Volume of solution `=(100)/(1.787)mL`
`=(100)/(1.787xx1000)"litre"`
`:. M_(H_(2)SO_(4))={(80)/(98xx(100)/(1.787xx1000))}=14.59`
Let `V mL` of this `H_(2)SO_(4)` are used to prepare `1 "litre"` of `0.2MH_(2)SO_(4)`, then `mM` of conc.
`H_(2)SO_(4)=mM` of dil. `H_(2)SO_(4)` (`mM` does not change on dilution)
`Vxx14.59= 1000xx0.2`
`V= 13.71 mL`