Question

What would be the Equivalent mass of $F{e}_{0.9}O$ in reaction with acidic $K_{2}C{r}_2{O}_7$? (M= Molar mass)
(A) ${\displaystyle \frac{7M}{10}}$
(B) ${\displaystyle \frac{10M}{7}}$
(C) ${\displaystyle \frac{7M}{9}}$
(D) $9M$

Answer 1

Hint: As we know that equivalent weight of an element or compound is that weight which is either reacted or displaced from $1g$ hydrogen, $8g$ oxygen or $35.5g$ chlorine.

Formula used: Equivalent weight $={\displaystyle \frac{molecularweight}{n-factor}}$

Complete step by step solution:
We can define the equivalent weight of an element or compound as that weight which is either reacted or displaced from $1g$ hydrogen, $8g$ oxygen or $35.5g$ chlorine and it is basically calculated using the formula:
$Equivalentweight={\displaystyle \frac{molecularweight}{n-factor}}$, where n-factor is calculated by change in oxidation state.
We know that Potassium dichromate is a strong oxidising agent in acidic medium. We can write the redox equation of this reaction as:
$6FeO+K_{2}C{r}_2{O}_7+13H_{2}S{O}_4\to 3F{e}_2(S{O}_4{)}_3+K_{2}S{O}_4+C{r}_2^{3+}+13H_{2}O$
Thus, when $F{e}_{0.9}O$ reacts with acidified potassium dichromate, the oxidation state of iron changes from $+2$ to $+3$. And iron in reactant is present in $0.9$ quantity, so the n-factor can be calculated here as:
$$n-factor=\left(3-{\displaystyle \frac{2}{0.9}}\right)\times 0.9=0.7$$
So the equivalent weight of iron can now be calculated using the above formula and we will get:
$Equivalentweight={\displaystyle \frac{molecularweight}{n-factor}}$
$Equivalentweight={\displaystyle \frac{M}{0.7}}$
$Equivalentweight={\displaystyle \frac{10\times M}{7}}$

Hence, the correct answer is option (A).

Note: Equivalent weight of acids and base can also be calculated using the same formula where n-factor is simply replaced by the acidity and basicity of the acid and base respectively. Basicity is the number of replaceable hydrogen present in acid and acidity is the number of hydroxide ions produced in solution by base.