A sample of `H_(2) SO_(4)` (density `1.787 g mL^(-1)`) si labelled as 86% by weight. What is the molarity of acid? What volume of acid has to be used to make `1 L` of `0.2 M H_(2) SO_(4)` ?

Explanation

`H_(2) SO_(4)` is 86% by weight
`:.` weight fo `H_(2) SO_(4) = 86 g`
Weight of solution `= 100 g`
`:.` Volume of solution `= (100)/(1.787) mL = (100)/(1.787 xx 1000) L`
`M_(H_(2)SO_(4)) = (86)/(98 xx (100)/(1.787 xx 1000)) = 15.68`
Let `V mL` of this `H_(2) SO_(4)` are used to prepare `1 L` of `0.2 M`
`H_(2) SO_(4)`
`:.` mmoles of concentrated `H_(2) SO_(4)` = mmoles of dilute `H_(2) SO_(4)`
`:. V xx 15.68 = 1000 xx 0.2`
`:. V = 12.75 mL`