Question

Equivalent mass of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is M/x, where the value of x is _____

Answer 1

Hint: For compounds that can undergo redox reaction we need to consider the reaction. Then by calculation the n factor we can determine the equivalent weight.

Complete step by step answer:
When the substance is not present in chemical reaction or present in chemical reactions (redox or non redox) the equivalent weight (E) is given as,
\[\begin{align}
  & E=\dfrac{\text{Molecular Mass}}{\text{n factor}}\text{ (for molecules)} \\
 & =\dfrac{\text{Atomic Mass}}{\text{n factor}}\text{ (for elements)} \\
 & =\dfrac{\text{Formula Mass}}{\text{n factor}}\text{ (for other species like radicals)} \\
\end{align}\]

Let us discuss the n factor or chemical equivalent.
For acid, n factor is nothing but the basicity ( for example n factor i.e. basicity of ${{H}_{2}}S{{O}_{4}}$ is 2) Similarly, for bases n factor is nothing but the acidity ( for example n factor for NaOH is 1)
For salts n factor is $\sum\limits_{{}}^{{}}{{{\text{n}}_{\oplus }}{{\text{v}}_{\oplus }}}$ ( where \[{{n}_{\oplus }}\] is the number of dissociated cations and \[{{v}_{\oplus }}\] is the valency )
Now for the given question we need to consider the dissociation of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
In acidic medium, the dissociation reaction of${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is
\[{{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\text{ + 14}{{\text{H}}^{\text{+ }}}\text{+ 6}{{\text{e}}^{\text{-}}}\to \text{ 2}{{\text{K}}^{\text{+}}}\text{ + 2C}{{\text{r}}^{\text{3+}}}\text{ + 7}{{\text{H}}_{\text{2}}}\text{O}\]

Here the number of electrons participating in the reduction is 6, therefore the chemical equivalent of the n factor is 6.
Again the molecular weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is 294
Hence, according to the definition, equivalent weight of ${{\text{K}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ is $\dfrac{294}{6}=49$
Therefore x is equal to 6

Note: Equivalent weight is related to the number of electrons released or accepted. So it can also be done by subtracting the value of lowest stable oxidation state from the highest stable oxidation state to get the n factor in this case.