What would be the equivalent conductivity of a cell in which 0.5 N
salt solution offers a resistance offers 40 ohm whose electrodes
ara 2 cm apart and 5cmº in area?
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Given:
Normality, N = 0.5 N
Resistance, R = 40 ohm
Length, l = 2 cm
Area, A = 5 cm²
To Find:
The equivalent conductivity of the cell.
Calculation:
- The conductivity of the cell is given by:
κ = l / AR
⇒ κ = 2 / (5 × 40)
⇒ κ = 0.01 S/cm
- The equivalent conductivity is given as:
Λeq = (κ × 1000) / N
⇒ Λeq = (0.01 × 1000) / 0.5
⇒ Λeq = 20 S cm²/ eq.
- So, the equivalent conductivity of the cell is 20 S cm²/ eq.
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