Question

# The molar conductance of NaCl, HCl and $C{{H}_{3}}COONa$ at infinite dilution is 126.45, 426.16 and 91 $oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$respectively. The molar conductance of $C{{H}_{3}}COOH$at infinite dilution is:[A] 201.28$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$[B] 390.71$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$[C] 698.28$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$[D] 540.48$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$

Hint: We can write the equation for dissociation of NaCl, HCl and$C{{H}_{3}}COONa$ where the sum of molar conductance of each ion will be equal to the respective compound it dissociated from. To find the conductance of$C{{H}_{3}}COOH$, we have to add or subtract the obtained equation as required to get an equation for molar conductance of the dissociated ions of $C{{H}_{3}}COOH$.

Complete step by step solution:
At a given constant temperature, the conductivity of solutions containing different electrolytes differs due to the concentration of the ions and difference in charge and size of the electrolytes when they dissociate into ions. Therefore, we needed to use the molar conductivity of each ion which is given as-
${{\lambda }_{m}}=\dfrac{k}{c}$
Where, ${{\lambda }_{m}}$is the molar conductance
K is the specific conductivity and
C is the concentration of the solution.
At infinite dilution, we can write the molar conductivity of NaCl as
$\lambda {{_{m}^{\infty }}_{NaCl}}=\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }$
Where, $\lambda {{_{m}^{\infty }}_{NaCl}}$is the molar conductivity of NaCl at infinite dilution and $\lambda _{N{{a}^{+}}}^{\circ }$and $\lambda _{C{{l}^{-}}}^{\circ }$is the conductance of sodium ion and chlorine ion respectively.
Similarly for HCl, we have-
$\lambda {{_{m}^{\infty }}_{HCl}}=\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }$
And for $C{{H}_{3}}COONa$we have,
$\lambda {{_{m}^{\infty }}_{C{{H}_{3}}COONa}}=\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }$
As we can see in the question that molar conductance of NaCl, HCl and $C{{H}_{3}}COONa$is given as 126.45, 426.16 and 91 $oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$respectively at infinite dilution.
Therefore, putting these values in the above equations we will get-
$\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }=126.45$\cdots (1) \lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ }=426.16$\cdots$ (2)
$\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }=91$$\cdots$ (3)
We know that $C{{H}_{3}}COOH$dissociates as-
$C{{H}_{3}}COOH=C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
Therefore, we can write the equation for molar conductance of $C{{H}_{3}}COOH$as-
$\lambda {{_{m}^{\infty }}_{C{{H}_{3}}COOH}}=\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ }$

We can calculate the conductance of the required ions from equation (1), (2) and (3) by adding the values in equation (3) and (2) and subtracting (1) from it.
Therefore, $\lambda {{_{m}^{\infty }}_{C{{H}_{3}}COOH}}=(\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ })-(\lambda _{{{H}^{+}}}^{\circ }+\lambda _{C{{l}^{-}}}^{\circ })+(\lambda _{N{{a}^{+}}}^{\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{\circ })$
Or,$\lambda {{_{m}^{\infty }}_{C{{H}_{3}}COOH}}$= (426.91+91) – 126.45 = 517.91-126.45 = 391.46$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$
Therefore, the molar conductance of $C{{H}_{3}}COOH$at infinite dilution is 391.46$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$
The option closest to the value we calculated is [B] 390.71$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$
Therefore, we can say that the correct answer is option [B] 390.71$oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$

ADDITIONAL INFORMATION: For weak electrolytes as they are not completely dissociated in the solution, the more dilute the solute is higher is the molar conductivity but the stronger electrolytes are not strongly dependent on the concentration of the solution because they will dissociate easily irrespective of the concentration of the solution. Therefore, there is a regular increase in molar conductivity on dilution for strong electrolytes.

Note: At infinite dilution, there is no inter ionic attraction as the ions are completely dissociated. It is important here to remember that as the molar conductivity is dependent on the concentration of the solution and at infinite dilution the number of ions present per volume decreases, hence molar conductivity increases.