Question

A metal is known to form fluoride $M{F}_2$.When 10A of electricity is passed through a molten salt for 330sec,1.95g of metal is deposited.Find the atomic weight of M.What will be the quantity of electricity required to deposit the same mass of Cu from $CuS{O}_4$

Answer 1

Equivalent weigh of metal= $\frac{I.t}{96500}$

$\Rightarrow \frac{1.95}{E}=\frac{10\times 330}{96500}$

$\Rightarrow E_{metal}=57.0$

Atomic weight of metal= $57.0\times 2=114.0$ (As metal is bivalent $M{F}_2$)

If $W_{Cu}=1.95g$ then

Equivalent of $Cu=\frac{I.t}{96500}$

$\Rightarrow \frac{1.95}{63.6/2}=\frac{I.t}{96500}$

$\Rightarrow I.t=Q=5917.45C$