A lead storage battery delivers a constant current of 0.50 A for 9650 s during discharge. Use F = 96500 C/mol, Pb = 207, S = 32, O = 16. The total mass of $PbSO X_{4}$ formed on both plates is

Answer: B

Charge

= 0.50 \times 9650 = 4825

C, so

n (e^{-}) = 4825/96500 = 0.05

mol. The cell reaction transfers 2 electrons and forms 2 mol

PbSO X_{4}

(one on each plate), so 1 mol

PbSO X_{4}

forms per mole of electrons. Molar mass of

PbSO X_{4}

= 207 + 32 + 64 = 303

g/mol, so the mass

= 0.05 \times 303 = 15.15