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Question

During the discharge of a lead storage battery the density of sulphuric acid fell from 1.294 to 1.139g mL1 . H2SO4 of density 1.294 g mL1 is 39% and that of density 1.139 g mL1 is 20% by weight. The battery holds 3.5 L of acid and the volume practically remains constant during the discharge. Calculate the number of Ampere hours for which the battery must have been used. The discharging reactions are:
Pb+SO24PbSO4+2e(anode)
PbO2+4H++SO24+2ePbSO4+2H2O(cathode)

A
190
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B
530
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C
265
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D
132.5
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Solution

The correct option is C 265
Pb+PbO2+4H++2SO242PbSO4+2H2O
Here normality and molarity of H2SO4 are equal.
After electrolysis,molarity of HSO4 will be,
=20×1.139×100098×100=2.325
The number of moles = 2.325×3.5=8.1375
Before electrolysis,molarity of HSO4 will be,
=39×1.294×100098×100=5.15
The number of moles = 5.15×3.5=18.025
Equivalents of H2SO4 used = 18.0258.1375=9.8875m=ZIt
I×t=9.8875×96500÷3600=265.04 ampere hr

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