An ion with mass number 56 contains 3 units of positive charge

Solution:

Let the number of electrons present in ion $\mathrm{A}^{3+}$ be $x$.

$\therefore$ Number of neutrons in it $=x+30.4 \%$ of $x=1.304 x$

Since the ion is tripositive,

$\Rightarrow$ Number of electrons in neutral atom $=x+3$

$\therefore$ Number of protons in neutral atom $=x+3$

Given,

Mass number of the ion $=56$

$\therefore(x+3)+(1.304 x)=56$

$2.304 x=53$

$x=\frac{53}{2.304}$

$x=23$

$\therefore$ Number of protons $=x+3=23+3=26$

$\therefore$ The symbol of the ion ${ }_{26}^{56} \mathrm{Fe}^{3+}$.