
Unit of Equivalent conductance is:
(a)- $oh{{m}^{-1}}c{{m}^{2}}{{(g-eq)}^{-1}}$
(b)- $ohm\text{ }cm\text{ }(g-eq)$
(c)- $ohm\text{ }c{{m}^{2}}{{(g-eq)}^{-1}}$
(d)- $oh{{m}^{-1}}cm\text{ }{{(g-eq)}^{-1}}$
Answer
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Hint: The equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte. The formula is:
${{\lambda }_{eq}}={{\kappa }_{c}}\text{ x V}$
This volume is equal to:
$V=\dfrac{1000}{normality}$
Specific conductivity has units $oh{{m}^{-1}}c{{m}^{-1}}$, and normality has units $\dfrac{g-eq}{c{{m}^{3}}}$.
Complete step by step answer:
We can define the equivalent conductance or equivalent conductivity as the conductance produced due to all the ions produced when one gram equivalent of the electrolyte is dissolved in V $c{{m}^{3}}$ of the solution, keeping the distance between the electrodes one cm and the area of the electrode is so large that all the solution is contained between the electrodes. Lambda is the representation of the equivalent conductance.
So, mathematically the equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte. This is written as:
${{\lambda }_{eq}}={{\kappa }_{c}}\text{ x V}$
Where ${{\kappa }_{c}}$ is the specific conductivity, and V is the volume of the solution having electrolyte.
In terms of concentration, the volume of the solution containing one gram equivalent will be:
$V=\dfrac{1000}{normality}$
So the overall formula of equivalent conductance will be:
${{\lambda }_{eq}}={{\kappa }_{c}}\text{ x }\dfrac{1000}{Normality}$
Now, for finding the units of the equivalent conductance, the units of specific conductance is $oh{{m}^{-1}}c{{m}^{-1}}$, and normality has units $\dfrac{g-eq}{c{{m}^{3}}}$. Putting these in the formula, we get:
${{\lambda }_{eq}}=oh{{m}^{-1}}c{{m}^{-1}}\text{ x }\dfrac{c{{m}^{3}}}{g-eq}$
${{\lambda }_{eq}}=oh{{m}^{-1}}c{{m}^{-1+3}}\text{ x (g-eq}{{\text{)}}^{-1}}$
${{\lambda }_{eq}}=oh{{m}^{-1}}c{{m}^{2}}\text{ x (g-eq}{{\text{)}}^{-1}}$
So the units of equivalents conductance will be $oh{{m}^{-1}}c{{m}^{2}}\text{ x (g-eq}{{\text{)}}^{-1}}$.
Therefore, the correct answer is option (a).
Note: It must be noted that the volume in the normality must be taken in $c{{m}^{3}}$ not in liters because mostly concentration is taken in liters. The other units of equivalent conductance are ${{\Omega }^{-1}}c{{m}^{2}}e{{q}^{-1}}$, $S\text{ c}{{\text{m}}^{2}}\text{ e}{{\text{q}}^{-1}}$, etc.
${{\lambda }_{eq}}={{\kappa }_{c}}\text{ x V}$
This volume is equal to:
$V=\dfrac{1000}{normality}$
Specific conductivity has units $oh{{m}^{-1}}c{{m}^{-1}}$, and normality has units $\dfrac{g-eq}{c{{m}^{3}}}$.
Complete step by step answer:
We can define the equivalent conductance or equivalent conductivity as the conductance produced due to all the ions produced when one gram equivalent of the electrolyte is dissolved in V $c{{m}^{3}}$ of the solution, keeping the distance between the electrodes one cm and the area of the electrode is so large that all the solution is contained between the electrodes. Lambda is the representation of the equivalent conductance.
So, mathematically the equivalent conductance is calculated by multiplying the specific conductivity and the volume of the solution having one gram equivalent of the electrolyte. This is written as:
${{\lambda }_{eq}}={{\kappa }_{c}}\text{ x V}$
Where ${{\kappa }_{c}}$ is the specific conductivity, and V is the volume of the solution having electrolyte.
In terms of concentration, the volume of the solution containing one gram equivalent will be:
$V=\dfrac{1000}{normality}$
So the overall formula of equivalent conductance will be:
${{\lambda }_{eq}}={{\kappa }_{c}}\text{ x }\dfrac{1000}{Normality}$
Now, for finding the units of the equivalent conductance, the units of specific conductance is $oh{{m}^{-1}}c{{m}^{-1}}$, and normality has units $\dfrac{g-eq}{c{{m}^{3}}}$. Putting these in the formula, we get:
${{\lambda }_{eq}}=oh{{m}^{-1}}c{{m}^{-1}}\text{ x }\dfrac{c{{m}^{3}}}{g-eq}$
${{\lambda }_{eq}}=oh{{m}^{-1}}c{{m}^{-1+3}}\text{ x (g-eq}{{\text{)}}^{-1}}$
${{\lambda }_{eq}}=oh{{m}^{-1}}c{{m}^{2}}\text{ x (g-eq}{{\text{)}}^{-1}}$
So the units of equivalents conductance will be $oh{{m}^{-1}}c{{m}^{2}}\text{ x (g-eq}{{\text{)}}^{-1}}$.
Therefore, the correct answer is option (a).
Note: It must be noted that the volume in the normality must be taken in $c{{m}^{3}}$ not in liters because mostly concentration is taken in liters. The other units of equivalent conductance are ${{\Omega }^{-1}}c{{m}^{2}}e{{q}^{-1}}$, $S\text{ c}{{\text{m}}^{2}}\text{ e}{{\text{q}}^{-1}}$, etc.
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