Home
Class 12
CHEMISTRY
Resistance of a conductvity cell filled ...

Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 `Omega`. The conductivity of this solution is 1.29 `Sm^(-1)`. Resistance of the same cell when filled with 0.02M of the same solution is `520 Omega`. the molar conductivity of 0.02M solution of the electrolyte will be:

A

`124xx10^(-4) S m^(2) mol^(-1)`

B

`1240xx10^(-4) Sm^(2)mol^(-1)`

C

`1.24xx10^(-4) S m^(2) mol^(-1)`

D

`12.4 xx10^(-4) sm^(2)mol^(-1)`

Text Solution

Verified by Experts

`C=0.1 M,R=100ohm,`
`k=1.29 ohm ^(-1)m ^(-1),Q k=1/Rxx1/atherefore 1/a=KR=1.29xx100 ohm,`
`C=0.02 M, R=520,`
`1/Rxx1/a=1/520xx129=0.248 ohm ^(-1)m^(-1)`
K decrease with dilution.
`therefore C=0.02` and not 0.2
Also `A=Kxx1/(M("in" m//L))`
`=Kxx1/(Mxx10^(-3)(m//m^(3)))=(0.248xx1)/(0.02xx10^(-3))=12.4xx10^(-3) Sm pI^(-1)=124xx10^(-4) S m^(2) m oI^(-1)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Resistance of 0.2 M solution of an electrolue is 50 Omega . The specific conductance of the solution is 1.4 S m^^(-1) . The resistance of 0.5 M solution of the same electrolyte is 280. Omega . The molar conducitivity of 0.5 M solution of the electrolyte is S m^2 "mol"^(-1) is.

The conductivity of 0.1 m KCl solution is 1.29sm^(-1) . If the resistance of the cell filled with 0.1 M KCl is 100 ohm. Calculate the cell constant.

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 .

Resistance of 0.2 M solution of an electrolyte is 50Omega . The specific conductance of the solution is 1.4 Sm^(-1) . The resistance of 0.5 M solution of the same electrolyte is :