A fully charged battery contains $500 M L$ of $5.0$ $M H_{2} S O_{4}$ What is the concentration of $H_{2} S O_{4}$ in the battery after $6.0 A$ of current drawn from the battery for $13.40 h ?$

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Solution

Amount of change flown
$= 6.0 \times 13.4 \times 60 \times 60$
$= 289440 C$
$= 3$ moles of $e^{-}$
$H_{2} S O_{4} ⇌ 2 H^{+} + S O_{4}^{2 -}$
$∴ 1.5$ moles of $H_{2} S O_{4}$ is dissociated.
Initial amount of $H_{2} S O_{4} = \frac{1}{2} \times 5 = 2.5$ mol
Final amount of $H_{2} S O_{4} = 1$ mol
$∴$ concentration of $H_{2} S O_{4}$
$= \frac{1 m o l}{500} \times 100$ m
$= 2 M$

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A fully charged battery contains 500ML of 5.0 MH2SO4 What is the concentration of H2SO4 in the battery after 6.0A of current drawn from the battery for 13.40h?

Amount of change flown =6.0×13.4×60×60=289440C=3 moles of e−H2SO4⇌2H++SO2−4∴1.5 moles of H2SO4 is dissociated.Initial amount of H2SO4=12×5=2.5 molFinal amount of H2SO4=1 mol∴ concentration of H2SO4=1mol500×100 m=2M

A fully charged battery contains $500 M L$ of $5.0$ $M H_{2} S O_{4}$ What is the concentration of $H_{2} S O_{4}$ in the battery after $6.0 A$ of current drawn from the battery for $13.40 h ?$