# During the discharge of lead storage battery the density of sulphuric acid fell from 1.294 to 1.139gcm-3. The sulphuric acid of density 1.294gcm-3 is 39% H2SO4by mass and that of density 1.139gcm-3 is 20% by mass. The battery holds 3.5L of acid and the volume remains practically constant duing discharge. Calculate the no. of amper-hours for which battery must have been used.

Reaction involved in Lead storage battery are

Hence, 2 mol of electron are transferred for the consumption of 2 mol of H2SO4
2 H2SO4 =  2 F = 2 x 96500 C

H2SO4 is 39%by mass
Mass of H2SO4 in 1 cm3 =  (1.294 x 39 )/100 g

Mass of H2SO4 in 3.5 L = (1.294 x (39/100) x 3.5 x 1000 g)
Moles of H2SO4 = Mass/ molar mass
Moles of H2SO4 in 3.5 L = (1.294 x (39/100) x 3.5 x 1000 g)/98  = 18.02 mol.

Now, Similarly moles of  H2SO4 in 3.5 L of density 1.139 g cm-3

Mass of H2SO4 in 3.5 L = (1.139 x (39/100) x 3.5 x 1000
Moles of H2SO4 in 3.5 L = (1.139 x (39/100) x 3.5 x 1000 g)/98  = 8.135 mol

Moles of H2SO4 consumed during discharge = 18.02 - 8.135  = 9.885 mol

2 moles of H2SO4 = 2 x 96500 C
9.885  mol of H2SO4 = (2 x 96500 x 9.885 )/2

Now , Since 3600 C = 1 amp-hour
Therefore,
[(2 x 96500 x 9.885 )/2 ] C  = [(2 x 96500 x 9.885 )/2 ] x [(1/3600)] amp-hour = 264.97 amp-hour

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