_{m≤ft({NH}₄ {OH})}^{0} is equal to \_\_\_\_

Correct Answer is : Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0_{m≤ft({NH}₄ {Cl})}^{0}+_{m({NaOH})}^{0}-_{m({NaCl})}^{0}Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0

_{m≤ft({NH}₄ {OH})}^{0}+_{m≤ft({NH}₄ {Cl})}^{0}-_{m({HCl})}^{0}

_{m≤ft({NH}₄ {Cl})}^{0}+_{m({NaOH})}^{0}-_{m({NaCl})}^{0}

_{m≤ft({NH}₄ {Cl})}^{0}+_{m({NaCl})}^{0}-_{m({NaOH})}^{0}

_{m({NaOH})}^{0}+_{m({NaCl})}^{0}-_{m≤ft({NH}₄ {Cl})}^{0}

_{m≤ft({NH}₄ {OH})}^{0} is equal to \_\_\_\_

Correct Answer is : Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0_{m≤ft({NH}₄ {Cl})}^{0}+_{m({NaOH})}^{0}-_{m({NaCl})}^{0}Λm(NH4Cl)0+Λm(NaOH)0−Λm(NaCl)0

Test Index

KCET 2026 Chemistry Solved Paper

Question : 31 of 60

Marks: +1, -0

\Lambda_{m\left(\mathrm{NH}_{4} \mathrm{OH}\right)}^{0}

is equal to

\_\_\_\_

$\Lambda_{m\left(\mathrm{NH}_{4} \mathrm{OH}\right)}^{0}+\Lambda_{m\left(\mathrm{NH}_{4} \mathrm{Cl}\right)}^{0}-\Lambda_{m(\mathrm{HCl})}^{0}$
$\Lambda_{m\left(\mathrm{NH}_{4} \mathrm{Cl}\right)}^{0}+\Lambda_{m(\mathrm{NaOH})}^{0}-\Lambda_{m(\mathrm{NaCl})}^{0}$
$\Lambda_{m\left(\mathrm{NH}_{4} \mathrm{Cl}\right)}^{0}+\Lambda_{m(\mathrm{NaCl})}^{0}-\Lambda_{m(\mathrm{NaOH})}^{0}$
$\Lambda_{m(\mathrm{NaOH})}^{0}+\Lambda_{m(\mathrm{NaCl})}^{0}-\Lambda_{m\left(\mathrm{NH}_{4} \mathrm{Cl}\right)}^{0}$

Validate

Solution:

Concept:

By Kohlrausch’s law, the limiting molar conductivity of a weak electrolyte equals the sum of the limiting ionic conductivities of its constituent ions.

Explanation:

For

\mathrm{NH}_4\mathrm{OH}

\Lambda_{m\left(\mathrm{NH}_4\mathrm{OH}\right)}^{0}=\lambda_{\mathrm{NH}_4^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}

Write Kohlrausch equations for the given strong electrolytes:

\Lambda_{m\left(\mathrm{NH}_4\mathrm{Cl}\right)}^{0}=\lambda_{\mathrm{NH}_4^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}

\Lambda_{m(\mathrm{NaOH})}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}

\Lambda_{m(\mathrm{NaCl})}^{0}=\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}

Now combine them as:

\Lambda_{m\left(\mathrm{NH}_4\mathrm{Cl}\right)}^{0}+\Lambda_{m(\mathrm{NaOH})}^{0}-\Lambda_{m(\mathrm{NaCl})}^{0}

=\left(\lambda_{\mathrm{NH}_4^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\right)+\left(\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}\right)-\left(\lambda_{\mathrm{Na}^{+}}^{0}+\lambda_{\mathrm{Cl}^{-}}^{0}\right)

The

\lambda_{\mathrm{Na}^{+}}^{0}

and

\lambda_{\mathrm{Cl}^{-}}^{0}

terms cancel out, giving:

\lambda_{\mathrm{NH}_4^{+}}^{0}+\lambda_{\mathrm{OH}^{-}}^{0}=\Lambda_{m\left(\mathrm{NH}_4\mathrm{OH}\right)}^{0}

Thus the required expression is obtained.

Answer:

Option B

\Lambda_{m\left(\mathrm{NH}_4\mathrm{Cl}\right)}^{0}+\Lambda_{m(\mathrm{NaOH})}^{0}-\Lambda_{m(\mathrm{NaCl})}^{0}

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