Home
Class 12
CHEMISTRY
The conductance of a 0.0015 M aqueous so...

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of `1 cm^2`. The conductance of this solution was found to be `5xx10^(-7)S`. The pH of the solution is 4. The value of limiting molar conductivity `overset(o)wedge_(m)` od this weak monobasic acid in aqueous solution is `Zxx10^(2)Scm^(2)mol^(-1)`. the value of Z is ____________.

Text Solution

Verified by Experts

The correct Answer is:
6
Promotional Banner

Similar Questions

Explore conceptually related problems

The conductance of 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1 cm^(2) . The conductance of this solution was found to be 5xx10^(-7) S . The pH of the solution is 4. Calculate the value of limiting molar conductivity.

A 0.1 M solution of monobasic acid at specific resistance of r ohms-cm, its molar conductivity is

The equivalent conductance of M//32 solution of a weak monobasic acid is 8.0 and at infinite dilution is 400 . The dissociation constant of this acid is :

Resistance of 0.2 M solution of an electrolyte is 50 Omega . The specific conductance of the solution is 1.3 S m^(-1) . If resistance of the 0.4 M solution of the same electrolyte is 260 Omega , its molar conductivity is .

0.5N solution of a salt placed between two Pt electrodes 2.0 cm apart and having area of cross-section 2.5 cm^(2) has resistance of 25 ohms. Calculate the conductance and cell constant.

The resistance of decinormal solution is found to be 2.5 xx 10^(3) Omega . The equivalent conductance of the solution is (cell constant = 1.25 cm^(–1) )

The conductivity of a 0.01 M solution of acetic acid at 298 k is 1.65xx10^(-4) S cm ^(-1) . Calculate molar conductivity (wedge_(m)) of the solution.

The conductivity of a 0.01 M solution of acetic acid at 298 k is 1.65xx10^(-4) S cm ^(-1) . Calculate molar conductivity (wedge_(m)) of the solution.

Specific conductance of 0.1M nitric acid is 6.3xx10^(-2)ohm^(-1)cm^(-1) . The molar conductance of the solution is:

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^(-1) . Calculate its molar conductivity.