Question

Calculate the amount of charge required to deposit $1.0$ g of sodium metal(molar mass $23.0$ g $mo{l}^{-1}$) from sodium ion.

• A. $2098$
• B. $96500$
• C. $193000$
• D. $4196$

Answer 1

The correct option is D $4196$

Moles of Na $=\frac{1}{23}=0.0435\text{moles}$

$N{a}^{+}+e^{-}\to Na$

It states that for $1F$ consumes one mole of $Na$ is liberated.

$\therefore$ For $0.043$ moles, $0.0435F$ is required.

1 Faraday $=96500\text{coulomb}$

0.043 F $=0.0435\times 96500$

$=4196\text{coulomb}$

Therefore, option D is correct.