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Question

Answers

(A) 2.3

(B) 4.6

(C) 9.2

(D) 18.4

Answer
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\[\text{Cell constant = }\frac{l}{A}=\frac{\text{Specific conductance}}{\text{Conductance}}\]

Let’s get some insight about equivalent conductance.

Equivalent conductance of an electrolyte is defined as the conducting power of all the ions produced by dissolving one gram- equivalent of an electrolyte in solution.

Equivalent conductance is expressed as ${{\Lambda }_{\varepsilon }}$ and its relation with specific conductance can be given by the following equation.

\[{{\Lambda }_{\varepsilon }}=\frac{\kappa \times 1000}{C}=\frac{\kappa \times 1000}{N}\]

Here C is the concentration in gram equivalent per litre.

- This term has been quite frequently used. Now it is replaced by molar conductance which is almost the same except that it doesn’t take into account the number of equivalents in cases where one of the ions is multivalent.

Instead, in the case of molar conductance, conductance is divided by the number of moles per unit volume rather than the number of equivalents per unit volume. In cases when both ions are monovalent, the two of the terms are interchangeable with each other.

Now, let us understand what the cell constant really is:

The electrodes are not exactly 1 unit apart and so they may not possess a surface area of 1 square unit. So, the measured resistance does not give the value of specific conductance of the solution.

We know that

\[\text{Cell constant = }\frac{l}{A}=\frac{\text{Specific conductance}}{\text{Conductance}}\]

Let us now apply this knowledge to the given question.

\[\text{Equivalent conductance (}\kappa \text{) =}\frac{\text{Cell constant}}{\text{Resistance}}\text{=}\frac{1.15}{250}\text{ Sc}{{\text{m}}^{-1}}\]

So, we can write that

\[{{\Lambda }_{eq}}=\frac{1.15\times 1000}{250\times 1}\]

\[{{\Lambda }_{eq}}=4.6oh{{m}^{-1}}c{{m}^{2}}equi{{v}^{-1}}\]