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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)`. 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`

Text Solution

Verified by Experts

The correct Answer is:
B
`Lambda_(m)^(@)(HCOOH)=lambda^(@)(H^(+))+lambda^(@)(HCOO^(-))=349.6+54.6S cm^(2) mol^(-)`
`=404.2 S cm^(2) mol^(-)`
`Lambda_(m)^(c )=46.1S cm^(2) mol^(-1)` (Given)
`alpha=(Lambda_(m)^(c ))/(Lambda_(m)^(@))=(46.1)/(404.2)=0.114`
`{:(HCOOH hArr HCOO^(-)+H^(+)),("Initial conc. C mol L"^(-1)),("Conc. at eqm."C(1-alpha)C alpha " " C alpha):}`
`K_(alpha)=(C alpha.C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha)=(0.025xx(0.114)^(2))/(1-0.114)=3.67xx10^(-4)`
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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) .

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

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