Answer
425.4k+ views
Hint: The conductance of a solution of different electrolytes varies with their concentration. To compare the conductance of different electrolytes. It is convenient to define a quantity which is called equivalent conductance.
Complete answer:
- The equivalent conductance can be defined as the net conductance of every ion that is produced from one gram equivalent of a given substance.
- If we consider two large parallel electrodes set 1 cm apart and the whole of the solution containing 1g equivalent of an electrolyte is placed between the electrodes. If V is the volume of the solution containing 1 g equivalent of an electrolyte. The equivalent conductivity is given as:
\[\lambda =kV\]
Where k is the specific conductance,
If C is concentration of the solution in $\left( g\text{ }equi\text{ }c{{m}^{-3}} \right)$, then we can write the relation of Volume to concentration, that is $volume=\dfrac{1}{concentration}$.
Hence, Then the equation is,
\[\lambda =k\times \dfrac{1}{C}\]
- $\lambda $ is never determined directly, but is calculated from its specific conductance and concentration.
- Specific conductance k is the reciprocal of specific resistance $\rho $ (it is the resistance offered by a material 1cm in length and having an area of cross section $1c{{m}^{2}}$). Specific resistance has unit ohm cm, specific conductance has unit of$oh{{m}^{-1}}\text{ }c{{m}^{-1}}$
-Hence we can see that the unit of equivalent conductance is found to be:
\[\lambda =k\times \dfrac{1}{C}\]
\[\begin{align}
& \dfrac{oh{{m}^{-1}}c{{m}^{-1}}}{g\text{ equiv c}{{\text{m}}^{3}}} \\
& =oh{{m}^{-1}}\text{ }g\text{ equi}{{\text{v}}^{-1}}\text{ c}{{\text{m}}^{2}}\text{ } \\
\end{align}\]
Or we can write it as $oh{{m}^{-1}}\text{ c}{{\text{m}}^{2}}\text{ g equi}{{\text{v}}^{-1}}$
- Hence we can conclude that the option (d) is the correct answer that is the equivalent conductivity has the unit $oh{{m}^{-1}}\text{ c}{{\text{m}}^{2}}\text{ g equi}{{\text{v}}^{-1}}$.
Additional information:
- Experimental measurement of a solution is reciprocal of the resistance, therefore, the experimental determination of the conductance of a solution involves the measurement of its resistance.
- We have seen that conductivity k is the reciprocal of resistivity, that is $\rho $that is:
\[\begin{align}
& k=\dfrac{1}{\rho } \\
& and\text{ }\rho \text{=R}\dfrac{a}{l} \\
& k=\dfrac{1}{R}\left( \dfrac{1}{a} \right) \\
& k=G\left( \dfrac{l}{a} \right) \\
\end{align}\]
Where G is the conductance of the cell, l is the distance of separation of two electrodes, and $\dfrac{l}{a}$ cell constant.
Note:
- We should not get confused in terms of specific and equivalent conductance. Specific conductance is denoted by symbol k and equivalent conductance is denoted by symbol $\lambda $
- We can see that$\lambda $ is never determined directly, but always calculated from its specific conductivity and concentration.
Complete answer:
- The equivalent conductance can be defined as the net conductance of every ion that is produced from one gram equivalent of a given substance.
- If we consider two large parallel electrodes set 1 cm apart and the whole of the solution containing 1g equivalent of an electrolyte is placed between the electrodes. If V is the volume of the solution containing 1 g equivalent of an electrolyte. The equivalent conductivity is given as:
\[\lambda =kV\]
Where k is the specific conductance,
If C is concentration of the solution in $\left( g\text{ }equi\text{ }c{{m}^{-3}} \right)$, then we can write the relation of Volume to concentration, that is $volume=\dfrac{1}{concentration}$.
Hence, Then the equation is,
\[\lambda =k\times \dfrac{1}{C}\]
- $\lambda $ is never determined directly, but is calculated from its specific conductance and concentration.
- Specific conductance k is the reciprocal of specific resistance $\rho $ (it is the resistance offered by a material 1cm in length and having an area of cross section $1c{{m}^{2}}$). Specific resistance has unit ohm cm, specific conductance has unit of$oh{{m}^{-1}}\text{ }c{{m}^{-1}}$
-Hence we can see that the unit of equivalent conductance is found to be:
\[\lambda =k\times \dfrac{1}{C}\]
\[\begin{align}
& \dfrac{oh{{m}^{-1}}c{{m}^{-1}}}{g\text{ equiv c}{{\text{m}}^{3}}} \\
& =oh{{m}^{-1}}\text{ }g\text{ equi}{{\text{v}}^{-1}}\text{ c}{{\text{m}}^{2}}\text{ } \\
\end{align}\]
Or we can write it as $oh{{m}^{-1}}\text{ c}{{\text{m}}^{2}}\text{ g equi}{{\text{v}}^{-1}}$
- Hence we can conclude that the option (d) is the correct answer that is the equivalent conductivity has the unit $oh{{m}^{-1}}\text{ c}{{\text{m}}^{2}}\text{ g equi}{{\text{v}}^{-1}}$.
Additional information:
- Experimental measurement of a solution is reciprocal of the resistance, therefore, the experimental determination of the conductance of a solution involves the measurement of its resistance.
- We have seen that conductivity k is the reciprocal of resistivity, that is $\rho $that is:
\[\begin{align}
& k=\dfrac{1}{\rho } \\
& and\text{ }\rho \text{=R}\dfrac{a}{l} \\
& k=\dfrac{1}{R}\left( \dfrac{1}{a} \right) \\
& k=G\left( \dfrac{l}{a} \right) \\
\end{align}\]
Where G is the conductance of the cell, l is the distance of separation of two electrodes, and $\dfrac{l}{a}$ cell constant.
Note:
- We should not get confused in terms of specific and equivalent conductance. Specific conductance is denoted by symbol k and equivalent conductance is denoted by symbol $\lambda $
- We can see that$\lambda $ is never determined directly, but always calculated from its specific conductivity and concentration.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)