Equivalent conductivity at infinite dilution for sodium -potassium oxalate $\left( {{{\left( {CO{O^ - }} \right)}_2}N{a^ + }{K^ + }} \right)$ is;
Given :Molar conductivities of oxalate is \[73.5\;Sc{m^2}mo{l^{ - 1}}\],${K^ + }$and $N{a^ + }$ ions at infinite dilution are $148.2,\;50.1\;Sc{m^2}mo{l^{ - 1}}$ respectively.
A. $271.8\;Sc{m^2}e{q^{ - 1}}$
B. $67.95\;Sc{m^2}e{q^{ - 1}}$
C. $543.6\,\,Sc{m^2}e{q^{ - 1}}$
D. $135.9\,\,Sc{m^2}e{q^{ - 1}}$

Hint: We know that, molar conductivity is the power of conductance of the ions which is produced by the dissolving an electrolytic solution of one mole. In other words, it is also known as the efficiency of an electrolyte.

Complete step by step answer
The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to charge and size of the ions in which they dissociate.
Whenever the dissociation of any electrolyte is complete then each ion will make their fixed contribution in the molar conductance value. And this doesn’t depend upon the nature of electrolyte.
This can be calculated by using Kohlrausch law which states that the value of molar conductance of any electrolyte is calculated by adding the contribution from its constituent ion.
Given that,
Molar conductivities of oxalate are $73.5\;Sc{m^2}mo{l^{ - 1}}$,
Molar conductivities of oxalate are \[{K^ + }\]$148.2\,Sc{m^2}mo{l^{ - 1}}$and
\[N{a^ + }\]ions at infinite dilution are $50.1\,Sc{m^2}mo{l^{ - 1}}$.
To find-: Equivalent conductivity at infinite dilution for sodium -potassium oxalate
We know that in sodium potassium oxalate we have a total of three ions namely oxalate ion, sodium ion and potassium ion.
From Kohlrausch law, we know that the value of molar conductance of any electrolyte is calculated by adding the contribution from its constituent ion. Therefore, it is calculated as shown below.
$
\gamma _M^\infty = \gamma _M^\infty \left( {CO{O^ - }} \right) + \gamma _M^\infty \left( {N{a^ + }} \right) + \gamma _M^\infty \\
 = \left( {148.2 + 50.1 + 73.5} \right)S\,c{m^2}mo{l^{ - 1}}\\
 = 271.8\,S\,c{m^2}mo{l^{ - 1}}
$

Hence, option (A) is correct.

Note:
Both the conductivity and molar conductivity have different values at different concentrations of electrolyte. While the value of conductivity always decreases with decrease in concentration. This is due to an increase in the number of ions because of which a unit volume will decrease.