Question

The electrical resistance of a column of 0.05 M solution of
length 50 cm and area of cross-section 0.625 cm- is 5 x 1000 ohm.
Calculate its resistivity, conductivity and molar conductivity​

Answer 1

Answer:

l = 50 cm

A= πr2 = 3.14 × (0.5)2 = 0.785 cm Cell constant = lA = 500.785= 63.694 cm-1

(i) Resistivity = RCell constant = 5.55×10363.694 = 87.135 Ω cm.

(ii) Conductivity = 1resistivity=187.135 = 1.148 × 10-2s cm-1

(iii) Molar conductivity ∧m = 103KM = 103×1.148×10-20.05 ∧m = 229.6 s cm2 mol-1

Explanation:

= πr² = 3.14 × (1/2 cm)² = 0.785 cm² ; l = 50 cm

Resistivity , ρ = R × A / l

           = 5.55 × 10³ ohm ×0.785 cm² / 50 cm = 87.135 ohm cm

Conductivity, κ = 1/ρ = 1 / 87.135 ohm cm = 0.1148 S cm⁻¹

Molar Conductivity, Λm = κ × 1000 / M = 0.01148 S cm⁻¹ × 1000 / 0.05 mol L⁻¹

    FINAL RESULT = 229.6 S cm² mol⁻¹

Hope it helps you !!