Question

A metal is known to form fluoride $M{F}_2$. When 10 A of electricity is passed through a molten salt for 330 sec., 1.95 g of metal is deposited. The atomic weight of M is__________.

Answer 1

A metal is known to form fluoride $M{F}_2$. When 10 A of electricity is passed through a molten salt for 330 sec., 1.95 g of metal is deposited. The atomic weight of M is 114.
Since the metal forms $M{F}_2$, its valence is +2.
Thus, for deposition of 1 mole of metal, 2 moles of electrons are required.
The number of moles of electrons passed $=\frac{I\times t}{F}=\frac{10A\times 330sec}{96500F/mol{e}^{-}}=0.034$ moles electrons.
They will deposit $\frac{0.034}{2}=0.017$ moles of emtal.
However 1.65 g of metal is deposited. Hence the atomic weight of the metal is $\frac{1.95\text{g}}{0.017\text{mol}}=114\text{g/mol}$
Hence, the atomic weight of metal is 114 g/mol.