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The molar conductivity of 0.025 mol L^(-...

The molar conductivity of 0.025 mol `L^(-1)` methanoic acid is 46.1 S `cm^(2) mol^(-1)` . Calculate its dissociation constant.
(Given `lambda_(2(H^+)^@)=349.6` S `cm^2 mol^(-1)` and `lambda_("HCOO"^-)^@=54.6 S Cm^2 mol^(-1)` )

A

`3.14xx10^(-3)`

B

`3.67xx10^(-4)`

C

`4.15xx10^(-4)`

D

`5.21xx10^(-3)`

Text Solution

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The correct Answer is:
B
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The molar conductivity of 0.25 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Calculate the degree of dissociation constant. Given : lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) and lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)

The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol L^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

Knowledge Check

  • The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm^(2) mol^(-1) . Its degree of dissociation (alpha) and dissociation constant. Given lambda^(@)(H^(+))=349.6 S cm^(-1) and lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

    A
    `K_(a)=3.67xx10^(4)alpha=0.214`
    B
    `K_(a)=3.67xx10^(-4)alpha=0.114`
    C
    `K_(a)=2.25xx10^(-4)alpha=0.150`
    D
    `K_(a)=2.25xx10^(-2)alpha=0.314`

  • 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)`
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    C
    `0.178xx10^(-5)` mol `L^(-1)`
    D
    `0.0178xx10^(-5)` mol `L^(-1)`

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