Question

At \(25^\circ C\), the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl\(_2\) are \(248\times 10^{-4}\), \(126\times 10^{-4}\) and \(280\times 10^{-4}\,S\,m^2\,mol^{-1}\) respectively. \(\lambda_m^\circ\) of \(Ba(OH)_2\) in \(S\,m^2\,mol^{-1}\) is

• A. \(52.4\times 10^{-4}\)
• B. \(524\times 10^{-4}\)
• C. \(402\times 10^{-4}\)
• D. \(262\times 10^{-4}\)

Hint:

Use Kohlrausch law and eliminate common ions by subtraction. Then substitute in required electrolyte expression.

Answer 1

The Correct Option is B

Step 1: Use Kohlrausch’s law.
\[ \Lambda^\circ = \lambda^\circ_+ + \lambda^\circ_- \]
Step 2: Write given conductances.
For NaOH:
\[ \Lambda^\circ(NaOH)=\lambda^\circ_{Na^+}+\lambda^\circ_{OH^-}=248 \]
For NaCl:
\[ \Lambda^\circ(NaCl)=\lambda^\circ_{Na^+}+\lambda^\circ_{Cl^-}=126 \]
For BaCl\(_2\):
\[ \Lambda^\circ(BaCl_2)=\lambda^\circ_{Ba^{2+}}+2\lambda^\circ_{Cl^-}=280 \]
(All values in \( \times 10^{-4}\)).
Step 3: Find \(\lambda^\circ_{OH^-}\).
Subtract NaCl from NaOH:
\[ (\lambda_{Na^+}+\lambda_{OH^-})-(\lambda_{Na^+}+\lambda_{Cl^-}) = 248-126 \]
\[ \lambda_{OH^-}-\lambda_{Cl^-}=122 \]
Step 4: Find \(\lambda^\circ_{Ba^{2+}}\).
From BaCl\(_2\):
\[ \lambda_{Ba^{2+}} = 280 - 2\lambda_{Cl^-} \]
Step 5: Find \(\Lambda^\circ(Ba(OH)_2)\).
\[ \Lambda^\circ(Ba(OH)_2)=\lambda_{Ba^{2+}}+2\lambda_{OH^-} \]
Substitute:
\[ = (280 - 2\lambda_{Cl^-}) + 2(\lambda_{Cl^-}+122) \]
\[ = 280 -2\lambda_{Cl^-} +2\lambda_{Cl^-}+244 \]
\[ = 524 \]
So:
\[ \Lambda^\circ(Ba(OH)_2)=524\times 10^{-4}\,S\,m^2\,mol^{-1} \]
Final Answer:
\[ \boxed{524\times 10^{-4}\,S\,m^2\,mol^{-1}} \]