Question

0.5N solution of a salt placed between two pt electrodes 2.0 cm apart having area of cross section 2.5cm² has resistance of 25 ohms .calculate equivalent conductance and cell constant​

Answer 1

Answer:

here is ur answer hope this helps you.

Answer 2

The cell constant and equivalent conductance is 0.8 cm⁻¹ and 64 Ω⁻¹ cm² (g equivalent)⁻¹ respectively.

Given,

Normality = 0.5 N

Distance between electrodes = 2.0 cm

Area of cross-section = 2.5 cm²

Resistance = 25 Ω

To Find,

Equivalent conductance and Cell constant

Solution,

Cell Constant is given as -

Cell constant = $\frac{l}{a}$

where l is the distance between electrodes = 2.0 cm

and a is the Area of cross-section = 2.5 cm²

Cell constant = $\frac{2}{2.5}$

Cell constant = 0.8 cm⁻¹

Equivalent conductance is defined as the net conductance of every ion created from each gram equivalent of a given substance.

It is given as,

Λ = $\frac{k*1000}{C}$

where κ is the specific conductivity

and C is the normality

κ = G * Cell constant

where 'G' is the conductance

G = $\frac{1}{R}$

where R is the resistance

G = $\frac{1}{25}$

G = 0.04 Ω⁻¹

κ = G * Cell constant

κ = 0.04 * 0.8

κ = 0.032 ℧ cm⁻¹

Λ = $\frac{k*1000}{C}$

Λ = $\frac{0.032*1000}{0.5}$

Λ = $\frac{32}{0.5}$

Λ = 64 Ω⁻¹ cm² (g equivalent)⁻¹

Thus, the cell constant and equivalent conductance is 0.8 cm⁻¹ and 64 Ω⁻¹ cm² (g equivalent)⁻¹ respectively.

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