Question

According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte $F{e}_2(S{O}_4{)}_3$ is given by:

• A. ${\lambda }_m^0\left(F{e}^{3+}\right)+{\lambda }_m^0\left(S{O}_4^{2-}\right)$
• B. $\frac{2{\lambda }_m^0\left(F{e}^{3+}\right)+3{\lambda }_m^0\left(S{O}_4^{2-}\right)}{6}$
• C. $2{\lambda }_m^0\left(F{e}^{3+}\right)+3{\lambda }_m^0\left(S{O}_4^{2-}\right)$
• D. $3{\lambda }_m^0\left(F{e}^{3+}\right)+2{\lambda }_m^0\left(S{O}_4^{2-}\right)$

Answer 1

The correct option is C $2{\lambda }_m^0\left(F{e}^{3+}\right)+3{\lambda }_m^0\left(S{O}_4^{2-}\right)$

Kohlrausch law:
At infinite dilution, when all the interionic effects disappear and dissociation of electrolyte is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it is associated.

At infinite dilution, the ion becomes totally independent of other ions. At this point, it contributes maximum towards the conductance of the solution. This value of molar conductance is known as limiting molar conductivity. It is also called as molar conductivity at infinite dilution. It is represented by ${\Lambda }_m^0$

Kohlrausch law of independent migration of ions:
${\Lambda }_m^0={\lambda }_{+}^0+{\lambda }_{-}^0$

${\lambda }_{+}^0\to \text{Limiting molar conductivities of cation}$
${\lambda }_{-}^0\to \text{Limiting molar conductivities of anion}$

In general,
${\Lambda }_m^0=v_{+}{\lambda }_{+}^0+v_{-}{\lambda }_{-}^0$

$v_{+}$ and $v_{-}$ are the number of cations and anions after dissociation of an electrolyte.

For $F{e}_2(S{O}_4{)}_3$

$F{e}_2(S{O}_4{)}_3\to 2F{e}^{3+}+3S{O}_4^{2-}$

$\therefore$
${\Lambda }_m^0\left(F{e}_{2}S{O}_4\right)=2{\lambda }_{+}^0\left(F{e}^{3+}\right)+3{\lambda }_{-}^0\left(S{O}_4^{2-}\right)$