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The conductivity of 0.001028M acetic aci...

The conductivity of `0.001028M` acetic acid is `4.95xx10^(-5)Scm^(-1)`. Calculate its dissociation constant if `Lambda_(m)^(0)` for acetic acid is `390.5Scm^(2)mol^(-1)`.

Text Solution

Verified by Experts

We know,
`Lambda_(m)=kxx(1000)/(M)`.....(i)
Given : `"k=4.95 x 10"^(-5)"S cm"^(-1),M=0.001028`
`therefore"From"(i)Lambda_(m)=4.95xx10^(-5)xx(1000)/(0.001028)`
`="48.15 ohm"^(-1)cm^(2)mol^(-1)`
Degree of dissociation,
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(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 conductivity of 0.001028 M acetic acid is 4.95xx10^(-5) S cm^(-1) . Calculate dissociation constant if wedge_(m)^(@) for acetic acid is 390.5 S cm^(2) mol ^(-1) .

Knowledge Check

  • The conductivity of 0.001028 mol L^(-1) acetic acid is 4.95xx10^(-5) S cm^(-1) . Calculate its dissociation constant, if Lambda_m^@ , for acetic acid is 390.5 S cm^(2) mol^(-1)

    A
    `1.78xx10^(-5)` mol `L^(-1)`
    B
    `1.87xx10^(-5)` mol `L^(-1)`
    C
    `0.178xx10^(-5)` mol `L^(-1)`
    D
    `0.0178xx10^(-5)` mol `L^(-1)`

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