The density and equivalent weight of a metal | vedclass

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(B) Given:Density of metal $= 10.5\, g\, cm^{-3}$Equivalent weight of metal $= 100$Current,$I = 3\, A$Area $= 80\, cm^2$,thickness $= 0.005\, mm = 5 \

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(B) Given:Density of metal $= 10.5\, g\, cm^{-3}$Equivalent weight of metal $= 100$Current,$I = 3\, A$Area $= 80\, cm^2$,thickness $= 0.005\, mm = 5 	imes 10^{-4}\, cm$According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{E 	imes I 	imes t}{96500}$,where $E$ is the equivalent weight.$W = \frac{100 	imes 3 	imes t}{96500} \quad \dots (i)$Also,the mass $W$ can be calculated as $W = 	ext{density} 	imes 	ext{volume} = 	ext{density} 	imes 	ext{area} 	imes 	ext{thickness}$.$W = 10.5\, g\, cm^{-3} 	imes 80\, cm^2 	imes 5 	imes 10^{-4}\, cm = 0.42\, g \quad \dots (ii)$Equating $(i)$ and $(ii)$:$\frac{300 	imes t}{96500} = 0.42$$t = \frac{0.42 	imes 96500}{300} = 135.1\, s \approx 135\, s$.

The density and equivalent weight of a metal are $10.5\, g\, cm^{-3}$ and $100$,respectively. The time required for a current of $3\, A$ to deposit a $0.005\, mm$ thick layer of the same metal on an area of $80\, cm^2$ is closest to $....\, s$

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