Question

Given that ,$A_M^\infty$ and $A_{eq.}^\infty$ are molar and equivalent conductivities at infinite dilution; $\lambda$ is ionic conductivity at infinite dilution, then for potash alum:
These are multiple correct questions.
A. $A_M^\infty$ (potash alum) $= 2 \times \lambda _{{K^ + }}^\infty + 2 \times \lambda _{A{l^{3 + }}}^\infty + 4 \times \lambda _{SO_4^{2 - }}^\infty$
B. $A_{eq.}^\infty$ (potash alum) $= \dfrac{1}{8} \times A_M^\infty$ (potash alum)
C. $A_M^\infty$ (potash alum) $= \dfrac{1}{4} \times \lambda _{{K^ + }}^\infty + \dfrac{1}{4} \times \lambda _{A{l^{3 + }}}^\infty + \dfrac{1}{4} \times \lambda _{SO_4^{2 - }}^\infty$
D. $A_M^\infty$ (potash alum) $= \lambda _{{K^ + }}^\infty + \lambda _{A{l^{3 + }}}^\infty + \lambda _{SO_4^{2 - }}^\infty$

Answer 1

The terms mentioned in the question molar, ionic and equivalent conductivities at infinite dilution should be clear to proceed further in the question. Here, infinite dilution means that the electrolyte is $100\%$ ionized. The formula of potash alum is $[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}]$.

Formula used: Molar conductivity,
$A_M^\infty \left( {XY} \right) = \lambda _{{X^ + }}^\infty + \lambda _{{Y^ - }}^\infty$
Equivalent conductivity,
$A_{eq.}^\infty (XY) = \lambda _{eq({X^ + })}^\infty + \lambda _{eq({Y^ - })}^\infty$
$A_{eq.}^\infty = \dfrac{{A_M^\infty }}{{ch\arg e}}$

Complete step by step answer:
First, we will try to understand the meaning of the terms given in the question. So, let’s start with the molar conductivity of $A_M^\infty$. It is defined as the sum of ionic conductivities of an individual chemical compound. Equivalent conductivity $A_{eq.}^\infty$ is defined as the sum of ionic equivalent conductivities of an individual compound.
First, we will find some terms which will be required to solve the question like the total charge positive or negative on the given ion. Here we have potash alum $[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}]$. So potash alum can be written in form of ions as given below,
$[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}] = 2{K^ + } + 2A{l^{3 + }} + 4SO_4^{2 - }$
Now we can easily calculate the total charge present on the potash alum as $2 \times 1\left( {{K^ + }} \right) + 2 \times 3\left( {A{l^{3 + }}} \right) = 8$
So there will be a $+ 8$ or $- 8$ charge on potash alum. So we can consider the total charge of $8$.
Using the relation between the molar and equivalent conductivities we can get the relation as
$A_{eq.}^\infty = \dfrac{{A_M^\infty }}{{ch\arg e}}$
$\Rightarrow A_{eq.}^\infty = \dfrac{{A_M^\infty }}{8} = \dfrac{1}{8} \times A_M^\infty$
$\Rightarrow A_{eq.}^\infty = \dfrac{1}{8} \times A_M^\infty$ (Potash alum)

So we got the one option given in our question as to the result.
Now we will calculate molar conductivity using the formula and ion equation as given below,
$A_M^\infty \left( {XY} \right) = \lambda _{{X^ + }}^\infty + \lambda _{{Y^ - }}^\infty$
So now for potash alum we have,
$[{K_2}S{O_4}.A{l_2}{\left( {S{O_4}} \right)_3}] = 2{K^ + } + 2A{l^{3 + }} + 4SO_4^{2 - }$
Now molar conductivity for potash alum can be written as,
$A_M^\infty \left( {[{K_2}S{O_4}.A{l_2}{{\left( {S{O_4}} \right)}_3}]} \right) = 2 \times \lambda _{eq.({K^ + })}^\infty + 2 \times \lambda _{eq.(A{l^{3 + }})}^\infty + 4 \times \lambda _{eq.(SO_4^{2 - })}^\infty$
Final result: after calculation, we got two results
$A_{eq.}^\infty$ (Potash alum) $= \dfrac{1}{8} \times A_M^\infty$ (Potash alum) and,
$A_M^\infty$ (Potash alum) $= 2 \times \lambda _{eq.({K^ + })}^\infty + 2 \times \lambda _{eq.(A{l^{3 + }})}^\infty + 4 \times \lambda _{eq.(SO_4^{2 - })}^\infty$

So, the correct answer is Option A,B.

Note: The molar conductivity is also known as limiting molar conductivity and it is noted that it decreases with the increase in concentration.
Potash alum is also called potassium aluminum sulfate. It is a double salt, not complex salt.
We can easily derive the relation between molar and equivalent conductivity by dividing their formulas.