Question

The equivalent conductance of a solution containing $2.54g$ of $CuS{O_4}$ per litre is $91.0{\Omega ^{ - 1}}c{m^{ - 2}}e{q^{ - 1}}$. Its conductivity would be
A.$1.45 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}$
B.$2.17 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}$
C.$2.91 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}$
D.$4 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}$

Answer 1

Hint: We have to calculate the conductivity using the equivalent conductance and normality of the solution. The normality of the solution is calculated using the grams of the copper sulfate, molar mass of copper sulfate and the number of equivalence of the solution. We can obtain the conductance by multiplying the equivalent conductance and normality of the solution.

Complete answer:
Given data contains,
The equivalent conductance of a solution is $91.0{\Omega ^{ - 1}}c{m^{ - 2}}e{q^{ - 1}}$.
Mass of copper sulfate per liter is $2.54g$.
We know that the equivalent conductance can be given by the formula,
${\Lambda _{eq}} = \dfrac{{\kappa \times 1000}}{N}$
Here,
${\Lambda _{eq}}$ represents the equivalent conductance
$\kappa $ represents the conductance
$N$ represents the normality of the solution
We can calculate the normality of the solution using the given mass and molar mass of copper sulfate and the number of equivalence of the solution.
We know that the molar mass of copper sulfate is $159g/mol$.
So, we can write the normality of the solution as $\dfrac{{\dfrac{{2.54}}{1}}}{{\dfrac{{159}}{2}}}$.
Let us now rearrange the expression of the equivalent conductance to get the conductance of the solution.
We can write the expression for calculating the conductance as,
$\kappa = {\Lambda _{eq}} \times \dfrac{N}{{1000}}$
Let us now substitute the values of equivalent conductance and normality in the expression to calculate the conductance.
$\kappa = {\Lambda _{eq}} \times \dfrac{N}{{1000}}$
Substituting the known values we get,
$\kappa = \left( {91{\Omega ^{ - 1}}c{m^2}e{q^{ - 1}}} \right) \times \dfrac{{2.54 \times 2}}{{159 \times 1000}}$
On simplifying we get,
$\kappa = 2.91 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}$
The conductance of the solution is $2.91 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}$.
Therefore,option (C) is correct.

Note:
 We can say that conductivity (or) specific conductivity is the ability of a solution to conduct electricity. We can give SI units of conductivity as S/m. The unit of molar conductivity is $S{m^2}mo{l^{ - 1}}$. We can determine the limiting molar conductivity of any electrolytes with the help of Kohlraush’s law.