Question

The resistance of a solution A is 50 ohms and that of solution B is 100 ohms, both solutions being taken in the same conductivity cell. If equal volumes of solution A and B are mixed, the resistance (in ohms) of the mixture using the same cell will be \[13x + 2\] . Then x is:
Assume that there is no in the degree of dissociation of A and B on mixing.
(write the value to the nearest integer)

Answer 1

Hint:To solve this question, we must first calculate the individual conductivities of the given solutions. Then we must calculate the contribution of the solutions to the conductivity of their mixture. Using these values, we must calculate the final conductivity of the mixture. Then using the relation between conductance, resistance, conductivity, and cell constant, we can calculate the value of the resistance.

Formula Used: 1. Conductance \[ = \dfrac{1}{{resis\tan ce}}\]
                            2. Conductivity = conductance x cell constant

Complete Step-by-Step Answer:
For the given solutions:
1.Solution A:
Conductivity = conductance x cell constant
 \[{k_a} = G \times {G^*} = \left( {\dfrac{1}{5}} \right) \times ({G^*})\]
2.Solution B:
Conductivity = conductance x cell constant
 \[{k_b} = G \times {G^*} = \left( {\dfrac{1}{{100}}} \right) \times ({G^*})\]
Since sloth solutions are used in the same cell, the value of the cell constant or \[{G^*}\] for both the solutions is the same.
Since equal volumes of both solution A and solution B are mixed, the contribution of both these solutions gets halved to the conductivity of the final solution. Hence, their individual contribution towards the conductivity of the mixture can be given as: \[\dfrac{{{k_a}}}{2}\] and \[\dfrac{{{k_b}}}{2}\] .
Hence, the conductivity of the mixture is represented as the sum of the individual conductivities of the constituent solutions. Hence, the conductivity is: \[\dfrac{{{k_a} + {k_b}}}{2}\]
Hence, for the given mixture, the conductivity can be given as:
Conductivity = conductance x cell constant
 \[\dfrac{{{k_a} + {k_b}}}{2} = \dfrac{1}{R} = \times {G^*}\]
 \[\dfrac{{\dfrac{{{G^*}}}{{50}} + \dfrac{{{G^*}}}{{100}}}}{2} = \dfrac{1}{R} \times {G^*}\]
 \[\dfrac{{3{G^*}}}{{200}} = \dfrac{{{G^*}}}{R}\]
 \[R = \dfrac{{200}}{3}\]
R = 66.67 ohm

Note: The contribution of the constituent solutions to their mixture to the value of the conductivity of the entire mixture, is directly proportional to the relative concentration of the given solution in the mixture.