During the discharge of a lead storage battery, the density of `40% H_(2)SO_(4)` by weight fell from `1.225 ` to `0.98 (` which is `20%` by weight `).` What is the change in molarities of `H_(2)SO_(4)`?

Answer

3

Explanation

Discharging reaction `:`

`M_(1)=(% by weight xx 10xxd)/(Mw_(2))=(40xx1.225xx10)/(98)=5M`
`M_(2)=(% by weight xx10xxd)/(Mw_(2))=(20xx10xx0.98)/(98)=2M`
Change is molarities `=M_(1)-M_(2)=52=3M`