In how much time will one litre of {H_2} be c | vedclass

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(A) Given:Volume of ${H_2}$ $(V)$ = $1\, L = 10^{-3}\, m^3$Density of ${H_2}$ $(\rho)$ = $0.09\, kg/m^3$Current $(I)$ = $5\, A$Electrochemical equival

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$30$

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(A) Given:Volume of ${H_2}$ $(V)$ = $1\, L = 10^{-3}\, m^3$Density of ${H_2}$ $(ho)$ = $0.09\, kg/m^3$Current $(I)$ = $5\, A$Electrochemical equivalent $(Z)$ = $1 	imes 10^{-8}\, kg/C$Step $1$: Calculate the mass $(m)$ of ${H_2}$ gas.$m = ho 	imes V = 0.09\, kg/m^3 	imes 10^{-3}\, m^3 = 9 	imes 10^{-5}\, kg$Step $2$: Use Faraday's law of electrolysis: $m = Z 	imes I 	imes t$$9 	imes 10^{-5} = (1 	imes 10^{-8}) 	imes 5 	imes t$$t = \frac{9 	imes 10^{-5}}{5 	imes 10^{-8}} = 1.8 	imes 10^3\, s$Step $3$: Convert time into minutes.$t = \frac{1800\, s}{60\, s/min} = 30\, min$

In how much time will one litre of ${H_2}$ be collected by a $5\, A$ current? (Given: $Z = 1 \times {10^{ - 8}}\,kg/C$ and density of ${H_2} = 0.09\,kg/{m^3}$)

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