Home
Class 12
CHEMISTRY
Resistance of a decimolar solution betwe...

Resistance of a decimolar solution between two electrodes 0.02 meter apart and 0.0004 `m^2` in area was found to be 50 ohm. Specific conductance (k0 is :

A

`0.1Scm^(2)`

B

`1Scm^(2)`

C

`10Scm^(2)`

D

`4xx10^(-4)Scm^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`G^(*)=(l)/(A)=(0.02)/(0.0004)=50m^(-1)`
`(1)/(R)=(1)/(50)`
Spacific conductance `G^(*)xx(1)/(R)=1Scm^(2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The resistance of decinormal solution of a salt occupying a volume between two platinum electrodes 1.80 cm apart and 5.4 cm^(2) area was found to be 32 ohm . Calculate k and wedge _(eq) .

One normal salt solution surrounding two platinum electrodes, 2.1 cm apart and 6.3 cm^(2) in area was found to offer a resistance of 50 ohm. Calculate the equivalent conductivity solution.

(a). The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms. Calculate the equilvalet conductivity of the solution if the electrodes in the cell were 2 cm apart and each has an area of 3.5 sq. cm. (b). The conductivity of 0.001028M acetic acid is 4.95xx10^(-5)Scm^(-1) . Calculate its dissociation constant if ^^_(m)^(@) for acetic acid is 390.5Scm^(2_mol^(-1)

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 conductance of a salt solution (AB) measured by two parallel electodes of area 100 cm^2 separated by 10cm was found to be 0.0001Omega^(-1) . If volume enclosed between two electrode contain 0.1 mole of salt, what is the molar conductivity (Scm^2mol^(-1)) of salt at same concentration:

The specific conductance of a 0.1 N KCl solution at 23^(@)C is 0.012 "ohm"^(-1)"cm"^(-1) . The resistance of the cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be

The specific conductance of a 0.1 N KCI solution at 25^@ C is 0.015 ohm^ (-1) cm^(-1) . The resistances of the cell containing the solution at the same temperature was found to be 60 Omega . The cell constant (in cm^(-1)) will be

The specific conductance (k) of a 0.1M NaOH solution is 0.0221 S cm^(-1) . On addition of an equal volume (V) of 0.10 M HCI solution, the value of specific conductance (k) falls to 0.0056 S cm^(-1) . On further addition of same volume (V) of 0.10 M HCI , the value of k rises to 0.017 S cm^(-1) . calculate equivalent conductivity (^^) for HCI in S cm^(2) mol^(-1) .

The resistance of 0.5M solution of an electrolyte in a cell was found to be 50 Omega . If the electrodes in the cell are 2.2 cm apart and have an area of 4.4 cm^(2) then the molar conductivity (in S m^(2) mol^(-1)) of the solution is

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100Omega . The conductivity of this solution is 1.29 S m^(-1) . Resistance of the same cell when filled with 0.2 M of the same solution is 520Omega . The molar conductivity of 0.2 M solution of the electrolyte will be :