0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm^-1, calculate the molar conductivity of NaOH solution.

Given:

Cell constant (b) = 0.367 cm^–1
Molar concentration (C) = 0.05 M
Resistance (R) = 31.6 Ω

To find: Molar conductivity (^)

Formulae:

a. Cell constant, (b) = k x R

b. Molar conductivity (∧)=1000K/C

solution:

Cell constant, b = k x R

0.367=k x 31.6

`k=0.367/31.6=11.61 xx10^-3 Omega^-1 cm^-1`

from formula b

`"Molar conductivity" (∧)=(1000K)/C`

`=(1000xx11.61xx10^-3)/0.05`

`=232.2 Omega^-1 cm^2mol^-1`

Is there an error in this question or solution?

Q 3.2 | 3 marks

0.05 M NaOH solution offered a resistance of 31.6 Ω in a conductivity cell at 298 K. If the cell constant of the cell is 0.367 cm^-1, calculate the molar conductivity of NaOH solution. - Chemistry