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Q. On passing current through $KCl$ solution, $19.5 \,g$ of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is

Electrochemistry

Solution:

$\frac{\text { Weight of } K}{\text { Weight of } A l}=\frac{\text { Eq. weight of } K}{\text { Eq. weight of } A l}$

$\frac{19.5}{w_{ Al }}=\frac{39}{27 / 3}$

$w_{ Al }=\frac{27}{3} \times \frac{19.5}{39}=4.5\, g$