Question

The limiting molar conductance for aqueous solution of CaCl\(_2\) at 298K is 271.6 S cm\(^2\) mol\(^{-1}\). If the limiting ionic conductance of Ca\(^{2+}\) ion at the same temperature is 119 S cm\(^2\) mol\(^{-1}\) what is the limiting ionic conductance of Cl\(^-\) ion?

• A. 152.6 S cm\(^2\) mol\(^{-1}\)
• B. 76.3 S cm\(^2\) mol\(^{-1}\)
• C. 135.8 S cm\(^2\) mol\(^{-1}\)
• D. 228.7 S cm\(^2\) mol\(^{-1}\)
• E. 114.35 S cm\(^2\) mol\(^{-1}\)

Hint:

A common mistake is forgetting the factor of 2 for the Chloride ions.
Always write the balanced dissociation equation first: \(CaCl_2 \to Ca^{2+} + 2Cl^-\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Kohlrausch's Law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its individual ions.

Step 2: Key Formula or Approach:
For an electrolyte \(A_x B_y\):
\[ \Lambda^0_m = x \lambda^0_+ + y \lambda^0_- \]
For CaCl\(_2\):
\[ \Lambda^0_m(CaCl_2) = \lambda^0(Ca^{2+}) + 2 \lambda^0(Cl^-) \]

Step 3: Detailed Explanation:
Given:
\(\Lambda^0_m(CaCl_2) = 271.6 \text{ S cm}^2 \text{ mol}^{-1}\)
\(\lambda^0(Ca^{2+}) = 119 \text{ S cm}^2 \text{ mol}^{-1}\)

Substitute the values into the formula:
\[ 271.6 = 119 + 2 \times \lambda^0(Cl^-) \]
Subtract 119 from both sides:
\[ 2 \times \lambda^0(Cl^-) = 271.6 - 119 = 152.6 \]
Divide by 2:
\[ \lambda^0(Cl^-) = \frac{152.6}{2} = 76.3 \text{ S cm}^2 \text{ mol}^{-1} \]

Step 4: Final Answer:
The limiting ionic conductance of Cl\(^-\) is 76.3 S cm\(^2\) mol\(^{-1}\).