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(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = Z \cdot q$,where $Z$ is the $E.C.E.$ and $q$ is the total charg

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$m / 300$

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(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = Z \cdot q$,where $Z$ is the $E.C.E.$ and $q$ is the total charge passed.The total charge $q$ is equal to the area under the current-time $(I-t)$ graph.The time is given in minutes,so we convert it to seconds: $t = 6 	ext{ min} = 6 	imes 60 	ext{ s} = 360 	ext{ s}$.The area under the graph consists of a triangle (from $t = 0$ to $2 	ext{ min}$) and a rectangle (from $t = 2 	ext{ min}$ to $6 	ext{ min}$).Area of triangle $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2 	imes 60 	ext{ s}) 	imes 1 	ext{ A} = 60 	ext{ C}$.Area of rectangle $= 	ext{length} 	imes 	ext{width} = (6 - 2) 	imes 60 	ext{ s} 	imes 1 	ext{ A} = 4 	imes 60 	ext{ C} = 240 	ext{ C}$.Total charge $q = 60 	ext{ C} + 240 	ext{ C} = 300 	ext{ C}$.Using the formula $m = Z \cdot q$,we get $Z = \frac{m}{q} = \frac{m}{300} 	ext{ g/C}$.

In a copper voltameter,the mass deposited in $6 \text{ min}$ is $m \text{ g}$. If the current-time graph for the voltameter is as shown in the figure,then the $E.C.E.$ of the copper is:

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