The molar conductivity of a $1.5\,{\text{M}}$ solution of an electrolyte is found to be ${\text{138}}{\text{.9}}\,{\text{Sc}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$. Calculate the conductivity of this solution.
Answer
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Hint: Molar conductivity is the conductivity of one molar of electrolyte. Molar conductivity is determined by dividing the conductivity of electrolytes with the concentration of electrolyte.
Formula used: $K\, = \dfrac{{{\lambda _m} \times 1000}}{C}$
Complete step by step answer:
The conductivity produced by dissolving one mole of an electrolyte in solution is known as molar conductivity. The conductivity is the conductance produced by ions of an electrolyte on dissolution per unit volume of solution.
The relation between conductivity, molar conductivity and concentration is given by the following formula:
$\Rightarrow K\, = \dfrac{{{\lambda _m} \times 1000}}{C}$
Where,
K is the molar conductivity.
${\lambda _m}$ is the conductivity.
C is the concertation.
Rearrange the above formula for conductivity as follows:
\[\Rightarrow{\lambda _m}\, = \dfrac{{K \times C}}{{1000\,}}\]
Substitute $1.5\,{\text{M}}$ for concentration of electrolyte solution and $\Rightarrow{\text{138}}{\text{.9}}\,{\text{Sc}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$ for molar conductivity.
$\Rightarrow{\lambda _m}\, = \dfrac{{{\text{138}}{\text{.9}}\,{\text{Sc}}\,{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}\, \times 1.5\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}}}{{1000\,{\text{c}}{{\text{m}}^3}{{\text{L}}^{ - 1}}}}$
$\Rightarrow{\lambda _m}\, = 0.208\,{\text{S}}{\text{c}}{{\text{m}}^{ - 1}}$
So, ${\lambda _m}\,$ conductivity is $0.208\,{\text{S}}\,{\text{c}}{{\text{m}}^{ - 1}}$.
Note the answer in two significant numbers.
Therefore, the conductivity of the given electrolyte solution is $0.21\,{\text{S}}{\text{c}}{{\text{m}}^{ - 1}}$.
Note: The unit of molar conductance is ${\text{S}}{\text{c}}\,{{\text{m}}^{ - 1}}$. Here, S is Siemens, that is the unit of conductance. Siemens is equal to reciprocal of ohm and it is also equal to mho. So, ${\text{S}}\,\, = \,{\text{oh}}{{\text{m}}^{ - 1}}\, = \,{\text{mho}}$. When the molar conductivity is divided by equivalent amount of electrolyte then the conductivity is known as an equivalent conductance. M is the unit of molarity which is equal to mol/L because molarity is defined as the mole of solute dissolved in per litter of the solution.
Formula used: $K\, = \dfrac{{{\lambda _m} \times 1000}}{C}$
Complete step by step answer:
The conductivity produced by dissolving one mole of an electrolyte in solution is known as molar conductivity. The conductivity is the conductance produced by ions of an electrolyte on dissolution per unit volume of solution.
The relation between conductivity, molar conductivity and concentration is given by the following formula:
$\Rightarrow K\, = \dfrac{{{\lambda _m} \times 1000}}{C}$
Where,
K is the molar conductivity.
${\lambda _m}$ is the conductivity.
C is the concertation.
Rearrange the above formula for conductivity as follows:
\[\Rightarrow{\lambda _m}\, = \dfrac{{K \times C}}{{1000\,}}\]
Substitute $1.5\,{\text{M}}$ for concentration of electrolyte solution and $\Rightarrow{\text{138}}{\text{.9}}\,{\text{Sc}}{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}$ for molar conductivity.
$\Rightarrow{\lambda _m}\, = \dfrac{{{\text{138}}{\text{.9}}\,{\text{Sc}}\,{{\text{m}}^2}{\text{mo}}{{\text{l}}^{ - 1}}\, \times 1.5\,{\text{mol}}\,\,{{\text{L}}^{ - 1}}}}{{1000\,{\text{c}}{{\text{m}}^3}{{\text{L}}^{ - 1}}}}$
$\Rightarrow{\lambda _m}\, = 0.208\,{\text{S}}{\text{c}}{{\text{m}}^{ - 1}}$
So, ${\lambda _m}\,$ conductivity is $0.208\,{\text{S}}\,{\text{c}}{{\text{m}}^{ - 1}}$.
Note the answer in two significant numbers.
Therefore, the conductivity of the given electrolyte solution is $0.21\,{\text{S}}{\text{c}}{{\text{m}}^{ - 1}}$.
Note: The unit of molar conductance is ${\text{S}}{\text{c}}\,{{\text{m}}^{ - 1}}$. Here, S is Siemens, that is the unit of conductance. Siemens is equal to reciprocal of ohm and it is also equal to mho. So, ${\text{S}}\,\, = \,{\text{oh}}{{\text{m}}^{ - 1}}\, = \,{\text{mho}}$. When the molar conductivity is divided by equivalent amount of electrolyte then the conductivity is known as an equivalent conductance. M is the unit of molarity which is equal to mol/L because molarity is defined as the mole of solute dissolved in per litter of the solution.
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