Question

Specific conductivity of ${\text{0}}{\text{.01 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution is $6 \times {10^{ - 3}}{\text{ S c}}{{\text{m}}^{ - 1}}$.its molar conductivity is
A) $1200{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$
B) $600{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$
C) $60{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$
D) $2400{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$

Answer 1

The conductance of a unit cube material is known as specific conductivity. The units of specific conductivity are ${\text{S c}}{{\text{m}}^{ - 1}}$ or ${\text{S }}{{\text{m}}^{ - 1}}$. The ratio of specific conductivity to the molar concentration of the dissolved electrolyte is known as its molar conductivity. The units of molar conductivity are ${\text{S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$ or ${\text{S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$.

Complete solution:
We are given ${\text{0}}{\text{.01 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution. The number of hydrogen ions i.e. ${{\text{H}}^ + }$ ions is 2.
Thus, we can calculate the molarity of ${\text{0}}{\text{.01 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ by dividing the normal concentration by the number of hydrogen ions. Thus,
${\text{Molar concentration}} = \dfrac{{0.01{\text{ N}}}}{2} = 0.005{\text{ M}}$
Thus, the molar concentration of ${\text{0}}{\text{.01 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution is $0.005{\text{ M}} = 0.005{\text{ mol }}{{\text{L}}^{ - 1}}$.
The equation to calculate the molar conductivity of a solution is as follows:
$\Lambda = \dfrac{k}{C}$
Where $\Lambda$ is the molar conductivity of the solution,
$k$ is the specific conductivity of the solution,
$C$ is the molar concentration of the solution.
Substitute $6 \times {10^{ - 3}}{\text{ S c}}{{\text{m}}^{ - 1}}$ for the specific conductivity of the solution, $0.005{\text{ mol }}{{\text{L}}^{ - 1}}$ for the molarity of the solution. Thus,
$\Lambda = \dfrac{{6 \times {{10}^{ - 3}}{\text{ S c}}{{\text{m}}^{ - 1}}}}{{0.005{\text{ mol }}{{\text{L}}^{ - 1}}}}$
But ${\text{1 L}} = {10^3}{\text{ c}}{{\text{m}}^3}$. Thus,
$\Lambda = \dfrac{{6 \times {{10}^{ - 3}}{\text{ S c}}{{\text{m}}^{ - 1}}}}{{0.005{\text{ mol }}{{\text{L}}^{ - 1}}}} \times \dfrac{{{{10}^3}{\text{ c}}{{\text{m}}^3}}}{{\text{L}}}$
$\Lambda = 1200{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$
Thus, the molar conductivity of solution of ${\text{0}}{\text{.01 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution is $1200{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$.

Thus, the correct option is (A) $1200{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$.

Note: The conducting power of all the ions that are produced by dissolving one mole of any electrolyte in the solution is known as the molar conductivity of the solution. As the temperature of the solution increases, the molar conductivity of the solution increases. This is because as the temperature increases the interaction and the mobility of the ions in the solution increases. The molar conductivity of both strong and weak electrolytes increases as the dilution increases. This is because dilution increases the degree of dissociation and the total number of ions that carry current increases.