Question

Equivalent conductance and molar conductance of $\text{F}{{\text{e}}_{2}}{{(S{{O}_{4}})}_{3}}$are related by relation:
\[\begin{align}
 & \text{A}\text{. }{{\wedge }_{\text{eq}}}={{\wedge }_{\text{m}}} \\
 & \text{B}\text{. }{{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{3} \\
 & \text{C}\text{. }{{\wedge }_{\text{eq}}}=3{{\wedge }_{\text{m}}} \\
 & \text{D}\text{. }{{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{6} \\
\end{align}\]

Answer 1

Hint: The relation between equivalent conductance and molar conductance:
${{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{({{\text{n}}^{+}}\times {{\text{z}}^{+}})}$, where ${{\text{n}}^{+}}$ is the number of cations and ${{\text{z}}^{+}}$ is the charge on a cation.

Complete answer:
Equivalent conductance (${{\wedge }_{\text{eq}}}$) of the electrolyte is the conductance of a volume of solution containing one equivalent mass of the dissolved substance when placed between the two parallel electrodes, when 1 cm apart.

${{\wedge }_{\text{eq}}}=\dfrac{\text{K}\times 1000}{\text{Normality}}$
Molar conductivity (${{\wedge }_{\text{m}}}$) is the conductivity produced by dissolving 1 gram-mole or 1 mole of an electrolyte when placed between two large electrodes at 1cm apart.
${{\wedge }_{\text{m}}}=\dfrac{\text{K}\times 1000}{\text{Molarity}}$
-$\text{F}{{\text{e}}_{2}}{{(S{{O}_{4}})}_{3}}$ has two cations $\text{F}{{\text{e}}^{+3}}$ and three anions $\text{S}{{\text{O}}_{4}}^{2-}$. The charge on $\text{F}{{\text{e}}^{+3}}$ is +3.

Using the relation, ${{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{({{\text{n}}^{+}}\times {{\text{z}}^{+}})}$, in this ${{\text{n}}^{+}}$ (number of cations) is 2 and ${{\text{z}}^{+}}$(charge on each cation) is 3. So, the equivalent conductance is ${{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{(2\times 3)}$ , which becomes ${{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{6}$. Thus, the correct option is option ‘d’. [${{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{6}$].
So, the correct answer is “Option D”.

Additional Information:
Use of $\text{F}{{\text{e}}_{2}}{{(S{{O}_{4}})}_{3}}$: Solution is used in dyeing, and as a coagulant for industrial wastes. It is widely used in pigments, and in pickling baths for steel and aluminium.

Note: The charge on the cation and anion needs to be found out correct by the valency method. Moreover, the formula should be correctly remembered. The formula is ${{\wedge }_{\text{eq}}}=\dfrac{{{\wedge }_{\text{m}}}}{({{\text{n}}^{+}}\times {{\text{z}}^{+}})}$ not this ${{\wedge }_{\text{m}}}=\dfrac{{{\wedge }_{\text{eq}}}}{({{\text{n}}^{+}}\times {{\text{z}}^{+}})}$ . Do not replace ${{\wedge }_{\text{eq}}}$with${{\wedge }_{\text{m}}}$ .