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During the discharge of a lead storage b...

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 charge in molarities of `H_(2)SO_(4)` ?

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The correct Answer is:
3
Discharging reaction : `PbO_(2) + 4H^(oplus) + SO_(4)^(2-) + 2e^(-) hArr PbSO_(4) + 2H_(2)O`
`M_(1) = ("% by weight" xx 10 xx d)/(Mw_(2)) = (40 xx 1.225 xx 10)/(98)= 5M`
`M_(2)= ("% by weight " xx 10 xx d)/(98)= 2M`
Change in molarities `= M_(1) - M_(2) = 5-2=3M`
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