Question

The molar conductivities of KCl, NaCl and KNO\(_3\) are 152, 128 and 111 S cm\(^2\) mol\(^{-1}\) respectively. What is the molar conductivity of NaNO\(_3\)?

• A. 101 S cm\(^2\) mol\(^{-1}\)
• B. 87 S cm\(^2\) mol\(^{-1}\)
• C. -101 S cm\(^2\) mol\(^{-1}\)
• D. -391 S cm\(^2\) mol\(^{-1}\)

Hint:

Kohlrausch’s law allows ionic conductivities to be added/subtracted to find unknown molar conductivity.

Answer 1

The Correct Option is B

Step 1: Use Kohlrausch’s law of independent ionic migration.
\[ \Lambda_m(KCl) = \lambda_{K^+} + \lambda_{Cl^-} \]
\[ \Lambda_m(NaCl) = \lambda_{Na^+} + \lambda_{Cl^-} \]
\[ \Lambda_m(KNO_3) = \lambda_{K^+} + \lambda_{NO_3^-} \]
Step 2: Write required expression for NaNO\(_3\).
\[ \Lambda_m(NaNO_3) = \lambda_{Na^+} + \lambda_{NO_3^-} \]
Step 3: Eliminate unknown ionic conductivities.
Add NaCl and KNO\(_3\), then subtract KCl:
\[ \Lambda_m(NaNO_3) = \Lambda_m(NaCl) + \Lambda_m(KNO_3) - \Lambda_m(KCl) \]
Step 4: Substitute values.
\[ \Lambda_m(NaNO_3) = 128 + 111 - 152 = 239 - 152 = 87 \]
Final Answer:
\[ \boxed{87 \, S\,cm^2\,mol^{-1}} \]