2.644 g of metal (M) was deposited when 8040 coulombs of electricity was passed through molten…
2.644 g of metal (M) was deposited when 8040 coulombs of electricity was passed through molten $\mathrm{MF}_2$ salt. What is the atomic mass of $\mathrm{M}\left(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right)$
$63.47 u$
$65.54 u$
$31.74 u$
$61.48 u$
Solution
$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{M}$
8040 coulombs deposited $=2.644 \mathrm{~g}$ of metal
$\begin{aligned}
& 2 \times 96500 \text { coulombs will deposit }=\frac{2.644}{8040} \times 2 \times 96500 \\
& =63.46 \mathrm{~g}
\end{aligned}$
Hence, atomic mass of metal is $63.46 \mathrm{~g} \mathrm{~mol}^{-1}$.