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Molar conductivities (wedge(m)^(infty)) ...

Molar conductivities `(wedge_(m)^(infty))` at infinite dilution of `NaCl, HCl and CH_(3)COONa` are `126.4, 425.9 and 91.0 S cm^(2) mol^(–1)` respectively. `wedge_(m)^(infty)` for `CH_(3)COOH` will be:

A

`290.8 S cm^(2) mol^(–1) `

B

`390.5 S cm^(2) mol^(–1) `

C

`425.5 S cm^(2) mol^(–1) `

D

`180.5 S cm^(2) mol^(–1) `

Text Solution

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The correct Answer is:
C
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If the molar conductivities at infinite dilution of NaCl, HCl and CH_(3)COONa(NaAc) are 126.4, 425.9 and 91.0 S cm^(2)mol^(-1) respectively, what will be that of acetic acid (Hac)?

If the molar conductivities at infinited dilution of NaCl, HCl and CH_(3)COONa are 126.4, 426.1 and 91.0 ohm^(-1) cm^(2) mol^(-1) respectively. What will be the that of acetic acid?

Knowledge Check

  • Molar conductivities (Lambda_(m)^(@)) at infinite dilution of NaCl, HCl and CH_(3)COONa arc 126.4, 425.9 and 91.0 S cm^(2) mol^(-1) respectively. Lambda_(m)^(@) for CH_(3)COOH will be

    A
    `390.5 S cm^(2) mol^(-1)`
    B
    `425.5 S cm^(2) mol^(-1)`
    C
    `180.5 S cm^(2) mol^(-1)`
    D
    `290.8 S cm^(2) mol^(-1)`

  • Molar conductiveiy [Lambda _m] at infinite dilution of Na CL, HCl and CH_3 COONa are 126.4, 425, 9 and 91 0 S cm^2 "mol"^(-1) respectively, Lambda_m for CH_3COOH will be ,

    A
    ` 425.5 S cm^2 mol^(-1)`
    B
    ` 180 .5 S cm^2 "mol"^(-1)`
    C
    ` 290 ,8 S cm^2 "mol"^(-1)`
    D
    ` 390. 5 cm^2 "mol"^(-1)`

  • The limiting molar conductances wedge^(infty) for NaCl, KBr and KCl are 126, 152 and 150 S cm^(2) mol^(–1) respectively. The wedge^(infty) for NaBr will be:

    A
    302 S `cm^(2) mol^(–1)`
    B
    176 S `cm^(2) mol^(–1)`
    C
    278 S `cm^(2) mol^(–1)`
    D
    `128 S cm^(2) mol^(–1)`

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