Question

For the reaction $3Z{n}^{2+}+2K_4\left[Fe\right(CN{)}_6]\to K_{2}Z{n}_3[Fe(CN{)}_6{]}_2+6k^{+}$ what will be the equivalent weight of $K_4\left[Fe\right(CN{)}_6]$ is M ?

• A. M
• B. $\frac{M}{2}$
• C. $\frac{M}{3}$
• D. $\frac{M}{4}$

Answer 1

The correct option is C $\frac{M}{3}$

Given,

$3Z{n}^{2+}+2K_4\left[Fe\right(CN{)}_6]⟶K_{2}Z{n}_3[Fe(CN{)}_6{]}_2+6K^{+}$

From the above reaction it is clearly seen that $2$ moles of $K_4\left[Fe\right(CN{)}_6]$ loses $6K^{+}$ ions after the reaction.

$\Rightarrow 1$ mole of $K_4\left[Fe\right(CN{)}_6]$ will have cation loss of $3$.

$\therefore$ n factor=$3$

Now, Equivalent weight=$\frac{Molarmass}{nfactor}$

$\Rightarrow$ Equivalent weight of $K_4\left[Fe\right(CN{)}_6]$ here is=$\frac{M}{3}$