Let a fully charged lead-storage battery contains 1.5L of 5M H₂SO₄. What will be the concentration of H₂SO₄ in the battery after 2.5A current is drawn from the battery for 6 hour?

Number of moles of H₂SO₄ before electrolysis$=\frac{\mathrm{MV}}{1000}=\frac{5\times 1.5\times 1000}{1000}=7.5$
Amount of H₂SO₄ used,

$=\frac{1\mathrm{tE}}{96500}=\frac{2.5\times 6\times 3600\times 68}{96500}=54.839\mathrm{g}=\frac{54.839}{98}\mathrm{mol}=0.56\mathrm{mol}$
Remaining mol of H₂SO₄= 7.5 – 0.56 = 6.94
Remaining molarity $\frac{\mathrm{n}}{\mathrm{V}}\times 1000=\frac{6.94}{1.5\times 1000}\times 1000=4.626$