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The conductivity of$0\cdot 1\text{N}$$\text{NaOH}$solution is$0\cdot 022\text{ S c}{{\text{m}}^{-1}}$. To this solution an equal volume of $0\cdot 1\text{N}$HCl solution is added which results into decrease of the conductivity of solution to$0\cdot 0055\text{ S c}{{\text{m}}^{-1}}$. The equivalence of conductivity of $\text{NaCl}$solution is$\text{S c}{{\text{m}}^{2}}$ equi ${{\text{v}}^{-1}}$is:
A. $0\cdot 055$
B. $0\cdot 11$
C. 110
D. 50

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Hint: Conductance: The conductance is the property of the conductor (metallic as well as electrolytic) which facilitates flow of electricity through it.
Equivalence Conductivity: It is defined as the conductance of all the ions produced by one gram equivalent of an electrolyte in a given solution. It is denoted by$\vartriangle $ .

Formula used:
${{\vartriangle }_{\text{e}}}=\text{K}\times \dfrac{1000}{\text{C}}$ or$\text{K}\times \dfrac{1000}{\text{N}}$
Where K= conductivity
N= normality

Complete step by step answer:
In the statement given that normality of solution 0.1N whose conductance is $0\cdot 022\text{ S c}{{\text{m}}^{-1}}$& when $0\cdot 1\text{N}$ is added to solution, the conductance becomes equal to.
Equivalent conductance$=\text{K}\times \dfrac{1000}{\text{N}}$
Where \[\text{K}=0\cdot 055\]
 N$=0\cdot 1$
And NaOH and HCl give two electrons after reaction. So that equivalence conductance becomes double
Therefore,\[\vartriangle =2\times \dfrac{\text{K}\times \text{1000}}{\text{N}}\]
$\vartriangle =2\times \dfrac{0\cdot 055\times 1000}{0\cdot 1}$
On solving the equation we get equivalence conductance is$110\text{ S}-\text{c}{{\text{m}}^{2}}\text{ equi}{{\text{v}}^{-1}}$.

So, the correct answer is “Option C”.

Additional Information:
Molar conductance:- Molar conductance is defined as the capacity of the dissociate ions to conduct electricity when the solution is diluted. It is denoted by lambda$\left( \text{ }\!\!\lambda\!\!\text{ } \right)$and its unit is given by $\text{S c}{{\text{m}}^{2}}\text{ mo}{{\text{l}}^{-1}}$.
Mathematically it is expressed as
$\vartriangle \text{m}=\text{K/C}$
Where K is the specific conductivity and c is the concentration in mole per liter

Note:
noted that in equivalent conductance normality of solution taken whereas in molar conductance molarity of solution taken.
Furthermore normality of solution is defined as number of gram equivalents of solute present per liter of solution. It is denoted by ‘N’.
Normality (N) = number of gram equivalents of solute / volume of solution in liters
Whereas molarity of solution is defined as number of moles of the solute per liter
Molarity (M) = number of moles of solute / volume of solution in liters.