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

Question

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 (A) 4.5 g (B) 9.0 g (C) 13.5 g (D) 27 g

Tardigrade Admin · Accepted Answer

( text Weight of K/ text Weight of A l)=( text Eq. weight of K/ text Eq. weight of A l) (19.5/w Al )=(39/27 / 3) w Al =(27/3) × (19.5/39)=4.5 g

Q. On passing current through $K Cl$ 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

$4.5 g$

$9.0 g$

$13.5 g$

$27 g$

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

$\frac{19.5}{w _{A l}} = \frac{39}{27/3}$

$w_{A l} = \frac{27}{3} \times \frac{19.5}{39} = 4.5 g$

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

4.5g

9.0g

13.5g

27g