Question

A conductivity cell filled with 0.02M AgNO₃ gives at 25°C a resistance of 947 ohms. If the cell constant is 2.3 cm⁻¹ what is the molar conductivity of 0.02M AgNO₃ at 25°C?

Answer 1

0(NH4Cl) = λ NH +4 + λ Cl – of 0.02M AgNO3 at 25ºC

Given :

L0(NaOH) = λNa + + λ OH – 0

C = 0.02M,

R = 947Ω,

L0(NaCl) = λNa + + λ Cl – –1

b = 2.3 cm

Hence,

To find :

L0(NH4Cl) + L0(NaOH) – L0 (NaCl)

L =?

= λ NH +4 λ Cl – + λ Na + – λ OH – – λ Cl – – λ Na +

Solution :

= λ NH +4 + λ OH – –1

B = 2.3cm

b = k.R or k = Lº(NH4OH)

R = 947Ω

Thus, L0(NH4OH)

= 2.428 × 10–3Ω–1cm–1.

Therefore, 1000k

L = C =

Therefore, 1000(cm3 L–1 )× 2.428×10–3 (Ω –1cm–1 )

0.02(mol L–1 )

L = 121.4 Ω–1cm2mol–1

Therefore, L0(NH4Cl) + L0(NaOH) – L0(NaCl)

Lº(NH4OH) = 149.7(Ω–1cm2 mol–1) + 248.1(Ω–1cm2 mol–1) – 126.5 (Ω–1cm2 mol–1)

= 271.3 Ω–1cm2 mol–1

The molar conductivity of CH3OOH at zero

concentration = 271.3 Ω–1cm2 mol–1