Question

For a strong electrolyte, a plot of molar conductivity against \( (\text{concentration})^{1/2} \) is a straight line, with a negative slope, the correct unit for the slope is:

• A. S cm\(^2\) mol\(^{-3/2}\) L\(^{1/2}\)
• B. S cm\(^2\) mol\(^{-1}\) L\(^{1/2}\)
• C. S cm\(^2\) mol\(^{-3/2}\) L
• D. S cm\(^2\) mol\(^{-3/2}\) L\(^{-1/2}\)

Answer 1

The Correct Option is A

For a strong electrolyte, the molar conductivity \(\Lambda_m\) can be expressed as:
\[\Lambda_m = \Lambda_m^0 - A\sqrt{C}\]
where \(\Lambda_m^0\) is the molar conductivity at infinite dilution, \(A\) is a constant, and \(C\) is the concentration.
The term \(A\sqrt{C}\) has units of \(\text{S cm}^2 \text{mol}^{-1}\), so the units of \(A\) must be \(\text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}\) to ensure dimensional consistency when multiplied with \(\sqrt{C}\) (units of \(\text{mol}^{1/2} \text{L}^{-1/2}\)).

Answer 2

Step 1: Understanding the given information
For a strong electrolyte, molar conductivity (\( \Lambda_m \)) varies with the square root of concentration (\( c^{1/2} \)) according to the Kohlrausch’s law of independent migration of ions. The relationship is expressed as:
\[ \Lambda_m = \Lambda_m^0 - K \sqrt{c} \] where:
- \( \Lambda_m \) = molar conductivity at concentration \(c\),
- \( \Lambda_m^0 \) = molar conductivity at infinite dilution,
- \( K \) = slope of the line (a proportionality constant),
- \( c \) = concentration in mol L\(^{-1}\).

Step 2: Analyze the equation for the slope
From the above equation, if we plot \(\Lambda_m\) (in S cm² mol\(^{-1}\)) on the y-axis against \(c^{1/2}\) (in mol\(^{1/2}\) L\(^{-1/2}\)) on the x-axis, we get a straight line with:
- Intercept = \(\Lambda_m^0\)
- Slope = \(-K\)

Step 3: Determine the unit of the slope
To find the unit of slope \(K\), consider:
\[ \text{Unit of } K = \frac{\text{Unit of } \Lambda_m}{\text{Unit of } c^{1/2}} \] Now substitute the respective units:
\[ \text{Unit of } \Lambda_m = \text{S cm}^2 \text{mol}^{-1} \] and \[ \text{Unit of } c^{1/2} = (\text{mol L}^{-1})^{1/2} = \text{mol}^{1/2} \text{L}^{-1/2}. \] Hence, \[ \text{Unit of } K = \frac{\text{S cm}^2 \text{mol}^{-1}}{\text{mol}^{1/2} \text{L}^{-1/2}} = \text{S cm}^2 \text{mol}^{-3/2} \text{L}^{1/2}. \]

Step 4: Final conclusion
Therefore, the correct unit of the slope for the plot of molar conductivity vs. \((\text{concentration})^{1/2}\) is:

S cm² mol^−3/2 L^1/2