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(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = Z i t$.Since $i = \frac{V}{R}$,we have $m = \frac{Z V t}{R}$.Fo

Vedclass · Accepted Answer

$1.5$

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(B) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = Z i t$.Since $i = \frac{V}{R}$,we have $m = \frac{Z V t}{R}$.For the same voltameter,$Z$ and $R$ are constant,so $m \propto V t$.Therefore,$\frac{m_1}{m_2} = \frac{V_1 t_1}{V_2 t_2}$.Given $m_1 = 2 \ g$,$V_1 = 12 \ V$,$t_1 = 30 \ min$,$V_2 = 6 \ V$,and $t_2 = 45 \ min$.Substituting the values: $\frac{2}{m_2} = \frac{12 	imes 30}{6 	imes 45}$.$\frac{2}{m_2} = \frac{360}{270} = \frac{4}{3}$.$m_2 = 2 	imes \frac{3}{4} = 1.5 \ g$.

When a copper voltameter is connected with a battery of $e.m.f.$ $12 \ V$,$2 \ g$ of copper is deposited in $30 \ minutes$. If the same voltameter is connected across a $6 \ V$ battery,then the mass of copper deposited in $45 \ minutes$ would be ............. $g$.

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