Question

What fraction of $Fe$ exists as $Fe$ (III) in $Fe _{0.96} O$ ? (Consider $Fe _{0.96}$ to be made up of $Fe$ (II) and $Fe$ (III) only)

• A. $\frac{1}{12}$
• B. $0.08$
• C. $\frac{1}{16}$
• D. $\frac{1}{20}$

Answer 1

Answer: (A)

$Fe _{0.96} O$
Let, No. of ions of $Fe ^{+2}=x$
No. of ions of $Fe ^{+3}=0.96- x$
Now, $( x )(+2)+(0.96- x )(+3)+1(-2)=0$
$2 x+2.88-3 x-2=0$
$x =0.88$
$\therefore$ Fraction of $Fe ^{+3}$ ions $=\frac{0.08}{0.96}=\frac{1}{12}$.