Question

The limiting molar conductivities of the given electrolytes at 298 K follow the order: (${\lambda ^ \circ }({K^ + }) = 73.5$, ${\lambda ^ \circ }(C{l^ - }) = 76.3$, ${\lambda ^ \circ }(C{a^{ + 2}}) = 119.0$, ${\lambda ^ \circ }(S{O_4}^{ - 2}) = 160.0Sc{m^2}mo{l^{ - 1}}$).
(A) KCl < $CaC{l_2}$ < ${K_2}S{O_4}$
(B) KCl < ${K_2}S{O_4}$ < $CaC{l_2}$
(C) ${K_2}S{O_4}$ < $CaC{l_2}$ < KCl
(D) $CaC{l_2}$ < ${K_2}S{O_4}$ < KCl

Answer 1

Hint : Use the Kohlrausch’s Law and find the molar conductivities of KCl, $CaC{l_2}$and ${K_2}S{O_4}$by adding the conductivities of the ions present in the molecule. Then arrange them in increasing sequence of their conductivities.

Complete step by step solution :
-Kohlrausch’s Law:
$\lambda _{eq}^\infty = \lambda _c^\infty + \lambda _a^\infty$ (1) where, $\lambda _{eq}^\infty$ = equivalent conductivity or limiting molar conductivity of an electrolyte
$\lambda _c^\infty$ = equivalence conductivity of cation
$\lambda _a^\infty$ = equivalence conductivity of anion
-Here we will be using the Kohlrausch’s Law to find out the limiting molar conductivities of the given electrolytes.
The Kohlrausch’s Law states that the limiting molar conductivity of an electrolyte at infinite dilution is equal to the sum of conductances of the cations and anions.
Mathematically it is written as:
$\lambda _{eq}^\infty = \lambda _c^\infty + \lambda _a^\infty$ (1) where, $\lambda _{eq}^\infty$ = equivalent conductivity or limiting molar conductivity of an electrolyte
$\lambda _c^\infty$ = equivalence conductivity of cation
$\lambda _a^\infty$ = equivalence conductivity of anion
-So, now we will find the limiting molar conductivity of KCl, $CaC{l_2}$and ${K_2}S{O_4}$ using equation (1).

The question gives us the following values: ${\lambda ^ \circ }({K^ + }) = 73.5$, ${\lambda ^ \circ }(C{l^ - }) = 76.3$, ${\lambda ^ \circ }(C{a^{ + 2}}) = 119.0$, ${\lambda ^ \circ }(S{O_4}^{ - 2}) = 160.0Sc{m^2}mo{l^{ - 1}}$
Let us begin with KCl: It has 1 ${K^ + }$ ion and 1 $C{l^ - }$ ion, so its molar conductivity will be written as:

$\lambda _{KCl}^\infty = \lambda _{{K^ + }}^\infty + \lambda _{C{l^ - }}^\infty$
= 73.5 +76.3
= 149.8

Now for $CaC{l_2}$: Since it has 1 $C{a^{ + 2}}$ ion and 2 $C{l^ - }$ ions, it’s molar conductivity can be written as:
$\lambda _{CaC{l_2}}^\infty = \lambda _{C{a^{ + 2}}}^\infty + 2\lambda _{C{l^ - }}^\infty$
= 119 + 2(76.3)
= 271.6

For ${K_2}S{O_4}$: Since it has 2 ${K^ + }$ ions and 1 $SO_4^{ - 2}$ ion, its molar conductivity is written as:
$\lambda _{{K_2}S{O_4}}^\infty = 2\lambda _{{K^ + }}^\infty + \lambda _{S{O_4}^{ - 2}}^\infty$
= 2(73.5) + 160
= 307
-${K_2}S{O_4}$ has the highest limiting molar conductivity followed by $CaC{l_2}$ and then KCl.

So, the correct sequence is: (A) KCl < $CaC{l_2}$ < ${K_2}S{O_4}$

Note : While finding out the limiting molar conductivities of the electrolytes add the values of conductivity of cations and anions in their respective stoichiometric coefficients. Example for KCl it is: $\lambda _{KCl}^\infty = \lambda _{{K^ + }}^\infty + \lambda _{C{l^ - }}^\infty$ because there is only 1 ${K^ + }$ ion and 1 $C{l^ - }$ ion, but for $CaC{l_2}$ it is: $\lambda _{CaC{l_2}}^\infty = \lambda _{C{a^{ + 2}}}^\infty + 2\lambda _{C{l^ - }}^\infty$ because it has 1 $C{a^{ + 2}}$ ion and 2 $C{l^ - }$ ions.