Home
Class 12
CHEMISTRY
The conductivity of a 0.20 M KCl solutio...

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

Text Solution

Verified by Experts

Given, the conductivity is `kappa=0.0248S cm^(-1)` and the concentration is C = 0.20 M.
Then molar conductivity is calculated as,
`wedge_(m)=(kappaxx1000)/(M)`
`=(0.0248xx1000)//0.20`
`=124Scm^(2)mol^(-1)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The conductivity of 0.20 M solution of KCl at 298 K is 0.248S cm^(-1) . Calculate its molar conductivity.

Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar conductivity in this solution. If Lambda_(m)^(@) for acetic acid be 390.5 S cm^(2) "mol"^(-1) , what would be its dissociation constant? (i) First , find molar conductivity using the formula,, Lambda_(m) = ( K xx 1000)/( C ) (ii) Then find degree of dissociation ( alpha) and dissociation constant ( K_(a)) by using formula alpha = ( Lambda_(m))/( Lambda_(m)^(@)) and K_(a) = ( C alpha^(2))/( 1-alpha) respectively.

At a certain temperature the specific conductance of 0.40M KCl solution is 4.96xx10^-2 ohm^-1cm^-1 . Calculate its molar conductance.

Conductivity of 0.00241M acetic acid is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar condcutivity. If Lambda_(m)^(@) acetic acid is 390.5 S cm^(2) "mol"^(-1) , what is its dissociation constant?

Conductivity of 2.5 xx 10^(-4) M methonoic acid is 5.25 xx 10^(-5) S cm^(-1) . Calculate its molar Condcutivity and degree of dissociation. Given : lambda^(@) ( H^(+) = 349.5 S cm^(2) "mol"^(-1) and lambda^(@) ( HCOO^(-)) = 50.5 S cm^(2) "mol"^(-1) .

If specific conductivity of N// 50 KCl solution at 298 K is 0.002765 Omega^(-1) cm^(-1) and resistance of a cell contaning this solution is 100 Omega , calculate the cell constant.