Question

The resistance of 0.05M CH3COOH solution is 100 ohm. The cell constant is 0.035 cm^-1 .Calculate the molar conductivity of solution.

Answer 1

Answer:

We know

Conductivity= Conductance × cell constant= 1/Resistance × cell constant= 1/31.6 × 0.357 = 0.011 S

Molar conductivity= 1000× conductivity/ Molarity

= 1000× 0.011/0.05 = 220S cm² mol^-1

Answer 2

The molar conductivity of the solution is  7 Ω⁻¹ cm² mol⁻¹.

The resistance of 0.05 M CH₃COOH solution is 100Ω. The cell constant is 0.035 cm⁻¹.

We have to find the molar conductivity of solution.

First of all, we should find out electrolyte conductivity.

Electrolyte conductivity, k = cell constant/R

Here,

• cell constant = 0.035 cm⁻¹
• resistance, R = 100Ω

k = 0.035/100 = 3.5 × 10⁻⁴ Ω⁻¹ cm⁻¹

Now, molar conductivity is given by, C = (1000 × k)/M

Here,

• molarity, M = 0.05M
• electrolyte conductivity, k = 3.5 × 10⁻⁴ Ω⁻¹ cm⁻¹

C = (1000 ×  3.5 × 10⁻⁴ Ω⁻¹ cm⁻¹)/0.05 = 7 Ω⁻¹ cm² mol⁻¹

Therefore the molar conductivity of the solution is  7 Ω⁻¹ cm² mol⁻¹.

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