The electrical resistance of a column of 0.05 m naoh solution of diameter 1 cm and length 50 cm is 5.55 x 103 ohm. calculate its resistively, conductivity and molar conductivity.
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87
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
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
Answered by
133
Answer:
A = 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 !!
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