Question

The same quantity of electricity that liberated $2.158g$ silver was passed through a solution of a gold salt and $1.314g$ of gold was deposited. The equivalent mass of silver is $107.9$. Calculate the equivalent mass of gold. What is the oxidation state of gold in this gold salt? (At. mass of gold $=197$).

Answer 1

Given:-

Mass of SIlver$\left(m_1\right)=2.158g$

Mass of gold$\left(m_2\right)=1.314g$

Equivalent mass of $Ag\left(E_1\right)=107.9$

Equivalent mass of $Au\left(E_2\right)=?$

Formula:-

$\frac{m_1}{m_2}=\frac{E_1}{E_2}$

$\therefore \frac{2.158}{1.314}=\frac{107.9}{E_2}$

$\therefore E_2=\frac{107.9\times 1.314}{2.158}=65.7$

Equivalent weight of Gold$=\frac{Atomicmass}{Valency}$

$\therefore Valency=\frac{Atomicmass}{E_2}=\frac{197}{65.7}=3$

Oxidation state of Gold$=3$.