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Limiting molar conductivity of \left ( \text{i.e.}\ {\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{OH}) \right ) is equal to:

  • Option 1)

    {\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{Cl})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NaCl})-{\mathop\Lambda^{0}}_{\text{m}}(\text{NaOH})

  • Option 2)

    {\mathop\Lambda^{0}}_{\text{m}}(\text{NaOH})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NaCl})-{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{Cl})

  • Option 3)

    {\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{OH})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}{\text{Cl}})-{\mathop\Lambda^{0}}_{\text{m}}(\text{HCl})

  • Option 4)

    {\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{Cl})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NaOH})-{\mathop\Lambda^{0}}_{\text{m}}(\text{NaCl})

 

Answers (1)

best_answer

As we learnt in

Formula for limiting molar conductivity for electrolyte -

\Lambda_{m}^{0}= \lambda _{+}^{0}.\nu _{+}\:+\:\nu _{-}.\lambda_{-}^{0}

\lambda _{+}^{0}\:and\:\lambda_{-}^{0}

are limiting molar conductivities of cation and anion respectively.

- wherein

if an electrolyte on dissociation gives \nu_{+} \:cation\:and\: \nu_{-} anions.

 

 

 


Option 1)

{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{Cl})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NaCl})-{\mathop\Lambda^{0}}_{\text{m}}(\text{NaOH})

This option is incorrect.

Option 2)

{\mathop\Lambda^{0}}_{\text{m}}(\text{NaOH})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NaCl})-{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{Cl})

This option is incorrect.

Option 3)

{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{OH})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}{\text{Cl}})-{\mathop\Lambda^{0}}_{\text{m}}(\text{HCl})

This option is incorrect.

Option 4)

{\mathop\Lambda^{0}}_{\text{m}}(\text{NH}_{4}\text{Cl})+{\mathop\Lambda^{0}}_{\text{m}}(\text{NaOH})-{\mathop\Lambda^{0}}_{\text{m}}(\text{NaCl})

This option is correct.

Posted by

Aadil

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