
The molar conductivity of $ 0.1 $ M $ C{H_3}COOH $ is $ 4 $ S $ c{m^2}{\left( {mol} \right)^{ - 1}} $ . What is the specific conductivity and resistivity of the solution?
Answer
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Hint: The molar conductivity is defined as the conductivity of an electrolyte solution divided by molar concentration. The molar conductivity of a solution is obtained from molar concentration and conductivity. Specific conductivity will be represented by K, and molar concentration by M. The resistivity is the reciprocal of conductivity.
$ {\lambda _m} = \dfrac{K}{C} $
$ {\lambda _m} $ is molar conductivity
K is specific conductivity
C is molar concentration or molarity
Complete answer:
The molar conductivity is the conductance of all ions produced by one mole of a solution.
Given that molar conductivity is $ 4Sc{m^2}{\left( {mol} \right)^{ - 1}} $
The molar concentration of acetic acid is $ 0.1M $
Molar concentration is also known as molarity. It is defined as the number of moles of solute dissolved in one litre of the solution. It was represented by M.
Thus, the specific conductance can be determined by substituting the values of molar conductivity and molar concentration.
$ K = 4Sc{m^2}{\left( {mol} \right)^{ - 1}} \times 0.1M $
The $ 0.1M $ concentration will be equal to $ 0.1mol \times {10^{ - 3}}c{m^{ - 3}} $
By simplification,
$ K = 4Sc{m^2}{\left( {mol} \right)^{ - 1}} \times 0.1mol \times {10^{ - 3}}c{m^3} $
Thus, the specific conductance will be
$ K = 4 \times {10^{ - 4}}c{m^{ - 1}} $
This is also known as conductivity.
The conductivity is reciprocal to resistivity.
Thus, resistivity $ \rho = \dfrac{1}{K} $
By substituting the conductivity in the above equation, we will get
$ \rho = \dfrac{1}{{4 \times {{10}^{ - 4}}}} = 0.25 \times {10^4}cm $
Thus, the resistivity will be $ 0.25 \times {10^4}cm $
Note:
Resistivity is exactly reciprocal to conductivity. Conductivity is of different types like molar conductivity, equivalence conductivity and specific conductivity. All these conductivities can be determined from one another. Molar concentration is dependable on molarity or molar concentration.
$ {\lambda _m} = \dfrac{K}{C} $
$ {\lambda _m} $ is molar conductivity
K is specific conductivity
C is molar concentration or molarity
Complete answer:
The molar conductivity is the conductance of all ions produced by one mole of a solution.
Given that molar conductivity is $ 4Sc{m^2}{\left( {mol} \right)^{ - 1}} $
The molar concentration of acetic acid is $ 0.1M $
Molar concentration is also known as molarity. It is defined as the number of moles of solute dissolved in one litre of the solution. It was represented by M.
Thus, the specific conductance can be determined by substituting the values of molar conductivity and molar concentration.
$ K = 4Sc{m^2}{\left( {mol} \right)^{ - 1}} \times 0.1M $
The $ 0.1M $ concentration will be equal to $ 0.1mol \times {10^{ - 3}}c{m^{ - 3}} $
By simplification,
$ K = 4Sc{m^2}{\left( {mol} \right)^{ - 1}} \times 0.1mol \times {10^{ - 3}}c{m^3} $
Thus, the specific conductance will be
$ K = 4 \times {10^{ - 4}}c{m^{ - 1}} $
This is also known as conductivity.
The conductivity is reciprocal to resistivity.
Thus, resistivity $ \rho = \dfrac{1}{K} $
By substituting the conductivity in the above equation, we will get
$ \rho = \dfrac{1}{{4 \times {{10}^{ - 4}}}} = 0.25 \times {10^4}cm $
Thus, the resistivity will be $ 0.25 \times {10^4}cm $
Note:
Resistivity is exactly reciprocal to conductivity. Conductivity is of different types like molar conductivity, equivalence conductivity and specific conductivity. All these conductivities can be determined from one another. Molar concentration is dependable on molarity or molar concentration.
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