Question

Zinc can be determined volumetrically by the precipitation reaction as given below:
$3{Zn}^{2+}+2\left[K_{4}Fe{\left(CN\right)}_6\right]\to K_2{Zn}_3{\left[Fe{\left(CN\right)}_6\right]}_2+6K^{+}$
A sample of $Zn$ ore weighing $1.5432$ g was prepared for the reaction and it required $34.68$ mL of $0.1043M{K}_{4}Fe{\left(CN\right)}_6$ for titration. The mass percentage of zinc in the ore is (as nearest integer) :

Answer 1

Note the given reaction is not a redox change.

Also, mM of $Zn$ used $:$ mM of $K_{4}Fe{\left(CN\right)}_6$ used $::3:2$

$\therefore$ mM of $Zn$ used $=\frac{3}{2}$ mM of $K_{4}Fe{\left(CN\right)}_6$ used

$\frac{w}{65}\times 1000=\frac{3}{2}\times 34.68\times 0.1043$

$⟹$ $w=0.3527$ g

$\%$ purity of $Zn=$ $\frac{0.3527}{1.5432}\times 100=22.85$

So, the nearest integer is $23$.