Question

At 298K, the molar conductivities at infinite dilution(${{\lambda }_{m}}^{0}$) of $N{{H}_{4}}Cl$, $KOH$ and $KCl$ are 152.8, 272.6 and 149.8 $S\,c{{m}^{2}}\,mo{{l}^{-1}}$ respectively. The ${{\lambda }_{m}}^{0}$ of $N{{H}_{4}}OH$ is $S\,c{{m}^{2}}\,mo{{l}^{-1}}$ and percentage dissociation of 0.01 M $N{{H}_{4}}OH$ with ${{\lambda }_{m}}$= 25.15 $c{{m}^{2}}\,mo{{l}^{-1}}$ at the same temperature are:
A. 275.6, 0.91
B. 275.6, 9.1
C. 266.6, 9.6
D. 30, 84

Answer 1

Hint: Find out the molar conductivity of $N{{H}_{4}}Cl$ with the help of other compounds using the Kohlrausch’s Law and with the help of that, find the degree of dissociation.

Complete step-by-step answer:
In order to understand the solution, we need to know about Kohlrausch’s Law. According to Kohlrausch’s law, at infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductivity of electrolyte irrespective of the nature of the other ion with which it is related.
This implies that the molar conductivity of an electrolyte diluted infinitely can be expressed as the sum of the contributions from its specific ions. If ${{\lambda }_{+}}^{0}$ and ${{\lambda }_{-}}^{0}$ represent the limiting molar conductivities of cation and anion respectively, then the limiting molar conductivity of electrolyte at infinite dilution is given by:
\[{{\lambda }_{m}}^{0}={{v}_{+}}{{\lambda }_{+}}^{0}+{{v}_{-}}{{\lambda }_{-}}^{0}\]
Where v is the number of ions. Now, the ${{\lambda }^{0}}_{N{{H}_{4}}OH}$ can be written as (${{\lambda }^{0}}_{N{{H}_{4}}Cl}+{{\lambda }^{0}}_{KOH}+{{\lambda }^{0}}_{KCl}$).So,
${{\lambda }^{0}}_{N{{H}_{4}}Cl}=(152.8+272.6-149.8)S\,c{{m}^{2}}\,mo{{l}^{-1}}$
${{\lambda }^{0}}_{N{{H}_{4}}Cl}=275.6\,S\,c{{m}^{2}}\,mo{{l}^{-1}} $
We know, that degree of dissociation, which is denoted by $\alpha $ can be found out by the formula:
\[\alpha =\dfrac{{{\lambda }^{c}}}{{{\lambda }^{0}}}\]
Here, ${{\lambda }^{c}}$ and ${{\lambda }^{0}}$ represent molar conductivities at given concentration and at infinite dilution, respectively. So, by putting in the values, we get the value of degree of dissociation as:
\[\alpha =\dfrac{25.1}{275.6}=0.091\]
Now, in order to get the percentage dissociation, we multiply the result by 100, and it comes out to be 9.1.

So, we get option B as the correct answer for our question.

Note: It is to be noted that the Kohlrausch’s law is only applicable when it comes to weak electrolytes, because weak electrolytes can dissociate further on adding more concentration, whereas strong electrolytes will not dissociate much.