Advertisement Remove all ads

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

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.

Advertisement Remove all ads

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?
Advertisement Remove all ads

APPEARS IN

Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×