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If R(1) and R(2) are the resistance of ...

If `R_(1)` and `R_(2)` are the resistance of two solution of equal volume, then the conductance of the mixture in the same conductivity cell is:

A

`(1)/(2)((1)/(R_1) +(1)/(R_2))`

B

`(1)/(2)((1)/(R_(1)+R_(2)))`

C

`(1)/(sqrtR_(1) R_(2)`

D

none of these

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The correct Answer is:
To solve the problem regarding the conductance of a mixture of two solutions with equal volumes, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Conductance and Resistance**: - Conductance (G) is the reciprocal of resistance (R). Therefore, \( G = \frac{1}{R} \). - The specific conductance (κ) is defined as \( \kappa = \frac{1}{R} \cdot G^* \), where \( G^* \) is the cell constant. 2. **Calculate Specific Conductance for Each Solution**: - For the first solution, the specific conductance \( \kappa_1 \) can be expressed as: \[ \kappa_1 = \frac{G^*}{R_1} \] - For the second solution, the specific conductance \( \kappa_2 \) is: \[ \kappa_2 = \frac{G^*}{R_2} \] 3. **Mixing the Solutions**: - When equal volumes of the two solutions are mixed, the new specific conductance \( \kappa_{mix} \) for the mixture will be the average of the specific conductances of the two solutions, each halved due to dilution: \[ \kappa_{mix} = \frac{\kappa_1}{2} + \frac{\kappa_2}{2} \] 4. **Substituting the Values**: - Substitute the expressions for \( \kappa_1 \) and \( \kappa_2 \): \[ \kappa_{mix} = \frac{G^*}{2R_1} + \frac{G^*}{2R_2} \] - Factor out \( G^* \): \[ \kappa_{mix} = G^* \left( \frac{1}{2R_1} + \frac{1}{2R_2} \right) \] 5. **Finding the Conductance of the Mixture**: - The conductance \( G_{mix} \) can be expressed as: \[ G_{mix} = \kappa_{mix} \cdot \text{Cell Constant} \] - Since the cell constant remains the same, we can write: \[ G_{mix} = \frac{G^*}{2} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \] - This can be simplified to: \[ G_{mix} = \frac{G^*}{2} G_{net} \] - Where \( G_{net} = \frac{1}{R_1} + \frac{1}{R_2} \). 6. **Final Expression**: - Thus, the final expression for the conductance of the mixture is: \[ G_{mix} = \frac{G^*}{2} \left( \frac{1}{R_1} + \frac{1}{R_2} \right) \]

To solve the problem regarding the conductance of a mixture of two solutions with equal volumes, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Conductance and Resistance**: - Conductance (G) is the reciprocal of resistance (R). Therefore, \( G = \frac{1}{R} \). - The specific conductance (κ) is defined as \( \kappa = \frac{1}{R} \cdot G^* \), where \( G^* \) is the cell constant. ...
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