Home
Class 12
CHEMISTRY
The resistance of decinormal solution is...

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)`)

A

4.6

B

5.6

C

6.6

D

7.6

Text Solution

Verified by Experts

The correct Answer is:
A

Sp cond `="obs cond "xx " Cell constant"`
`=(1)/(2.5 xx 10^(3)) xx 1.5=4.6 xx 10^(-4)`
`wedge_(c)=(K xx 1000)/(N)=(4.6 xx 10^(-4) xx 1000)/(0.1)`
`4.6 Omega^(-1) cm^(-1)" eqv"`
Promotional Banner

Similar Questions

Explore conceptually related problems

The resistance of 0.1 N solution of a salt is found to be 2.5xx10^(3) Omega . The equivalent conductance of the solution is (Cell constant =1.15 cm^(-1) )

The resistance of 0.01N solution of an electrolyte AB at 328K is 100ohm. The specific conductance of solution is (cell constant = 1cm^(-1) )

A solution of 0.1 N KCl offers a resistance of 245 ohms. Calculate the specific conductance and the equivalent conductance of the solution if the cell constant is 0.571 cm^(-1) .

The resistance of a 1N solution of salt is 50 Omega . Calculate the equivalent conductance of the solution, if the two platinum electrodes in solution are 2.1 cm apart and each having an area of 4.2 cm^(2) .

Resistance of a 0.1 M KCl solution in a conductance cell is 300 ohm and specific conductance of "0.1 M KCl" is 1.33xx10^(-2)" ohm"^(-1)"cm"^(-1) . The resistance of 0.1 M NaCl solution in the same cell is 400 ohm. The equivalent conductance of the 0.1 M NaCl solution ("in ohm"^(-1)"cm"^(2)"/gmeq.") is

The resistance of a N//10 KCI solution is 245 ohms. Calculate the specific conductance and the equivalent conductance of the solution if the electrodes in the cell are 4 cm apart and each having an area of 7.0 sq cm.

If resistivity of 0.8M KCl solution is 2.5xx10^(3)Omega cm calculate lamda_(m) of the solution

The resistance of a 0.01N solution of an electrolyte was found to 210 ohm at 298K using a conductivity cell with a cell constant of 0.88 cm^(-1) . Calculate specific conductance and equivalent conductance of 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)

The resistance of a 0.01N solution of an electroyte was found to 210 ohm at 298K using a conductivity cell with a cell constant of 0.88 cm^(-1) . Calculate specific conductance and equilvalent conductance of solution.