Question

Molar conductance is used to define as the conducting power of all the ions produced by one mole of the electrolyte in the given solution.
$\Lambda m=\frac{1000\times K}{C}$
Where, K = specific conductance, C = Molar concentration of the solution.
At infinite dilution, each ion makes a definite contribution for the conductance of electrolyte. At infinite dilution, the molar conductance of an electrolyte is the sum of the ionic conductance of constituent ions.
$\Lambda A{B}_m^{\infty }=x{A}_{+}^{\infty }+y{A}_{+}^{\infty }$ where,x and y are the number of cations and anions per formula unit of the electrolyte. By measuring the conductance we can measure degree of dissociation and dissociation constants.
$\alpha =\frac{{\Lambda }^C}{{\Lambda }^{\infty }};K_a=\frac{a^{2}C}{(1-\alpha )};K_b=\frac{{\alpha }^{2}C}{(1-\alpha )}$
By conductance measurements we can calculate the solubility product of a sparingly soluble salt. For a sparingly soluble salt, the saturated solution will be an extremely dilute solution and $A^C$ is taken as $A^{\infty }$. Knowing $A^{\infty }$ (by Kohlrausch’s law) the concentration of saturated solution is calculated from which Ksp is calculated.
The value of ${\Lambda }^C$ for 0.001M aqueous solution of $N{H}_3$ is (in $S{m}^{2}mo{l}^{-2}$)

• A. $2.18\times {10}^{-2}$
• B. $3.01\times {10}^{-3}$
• C. $1.6\times {10}^{-5}$
• D. $1.17\times {10}^{-4}$

Answer 1

The correct option is B $3.01\times {10}^{-3}$

$K_b=\frac{a^{2}C}{(1-a)};a=\sqrt{\frac{K_b}{C}}=\sqrt{\frac{1.6\times {10}^{-5}}{0.001}}=0.1265$
$a=\frac{{\Lambda }^C}{{\Lambda }^{\infty }};{\Lambda }^C={\Lambda }^{\infty }\times \alpha =0.1265\times 2.3\times {10}^{-2}=3\times {10}^{-3}S{m}^{2}mo{l}^{-1}$