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The conducativity of water is 7.6 xx 10^...

The conducativity of water is `7.6 xx 10^(-2)Sm^(-1)` and the con-ductivity of `0.1 M` aqueous solution of `KCl` is `1.639 Am^(-1)`. A cell has a resistance of `33.20 Omega` when filled with `0.1 M Kcl` solution and `300 Omega` when filled with `0.1` M `CH_(3)CO_(2)CO_(2)H` solution. The molar conductivity of `CH_(3)CO_(2)CO_(2)H` is

A

`5.3 xx 10^(-4)Sm^(2)mol^(-1)`

B

`4.7 xx 10^(-4)Sm^(2)mol^(-1)`

C

`6.7 xx 10^(-4)Sm^(2)mol^(-1)`

D

`7.5 xx 10^(-4)Sm^(2)mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
First calculate conductivity due to `CH_(3)COOH`.
`kappa = K_("cell")//R`
i.e. `K prop 1//R`
Thus, we have
`(kappa(CH_(3)COOH))/(kappa(KCl)) = (R(KCl))/(R(CH_(3)COOH))`
or `kappa(CH_(3)COOH) = (R(KCl))/(R(CH_(3)COOH)) xx kappa(KCl)`
`= ((33.20 Omega)(1.1639 Sm^(-1)))/((300 Omega))`
`= 0.129 Sm^(-1)` Note that conductivity of sq. soln of `CH_(3)COOH` alos contains a contribution due to water. Thus, true value of conductivity of acetic acid will be
`kappa(CH_(3)COOH) = 0.129 S m^(-1) - 7.6 xx 10^(-2) S m^(-1)`
`= 0.053 s m^(-1)`
Finally
`Lambda_(m)(CH_(3)COOH) = (kappa)/(C) = (5.3 xx 10^(-2) S m^(-1))/(0.1 xx 10^(3) mol m^(-3))`
`= 5.3 xx 10^(-4)S m^(2) mol^(-1)`
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