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Question: The equivalent conductance of a solution containing \(2.54g\) of \(CuS{O_4}\) per litre is \(91.0{\O...

The equivalent conductance of a solution containing 2.54g2.54g of CuSO4CuS{O_4} per litre is 91.0Ω−1cm−2eq−191.0{\Omega ^{ - 1}}c{m^{ - 2}}e{q^{ - 1}}. Its conductivity would be
A.1.45×10−3Ω−1cm−11.45 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}
B.2.17×10−3Ω−1cm−12.17 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}
C.2.91×10−3Ω−1cm−12.91 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}
D.4×10−3Ω−1cm−14 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}

Explanation

Solution

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 91.0Ω−1cm−2eq−191.0{\Omega ^{ - 1}}c{m^{ - 2}}e{q^{ - 1}}.
Mass of copper sulfate per liter is 2.54g2.54g.
We know that the equivalent conductance can be given by the formula,
Λeq=κ×1000N{\Lambda _{eq}} = \dfrac{{\kappa \times 1000}}{N}
Here,
Λeq{\Lambda _{eq}} represents the equivalent conductance
κ\kappa represents the conductance
NN 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/mol159g/mol.
So, we can write the normality of the solution as 2.5411592\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,
κ=Λeq×N1000\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.
κ=Λeq×N1000\kappa = {\Lambda _{eq}} \times \dfrac{N}{{1000}}
Substituting the known values we get,
κ=(91Ω−1cm2eq−1)×2.54×2159×1000\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,
κ=2.91×10−3Ω−1cm−1\kappa = 2.91 \times {10^{ - 3}}{\Omega ^{ - 1}}c{m^{ - 1}}
The conductance of the solution is 2.91×10−3Ω−1cm−12.91 \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 Sm2mol−1S{m^2}mo{l^{ - 1}}. We can determine the limiting molar conductivity of any electrolytes with the help of Kohlraush’s law.