Question

The \(\Lambda_m\) (on y-axis) of \(NaCl\) and \(CsCl\) was plotted against \(\sqrt{c}\) \((c=\text{concentration on x-axis})\). Identify the correct figure for these electrolytes. \[ \Lambda_m=\Lambda_m^\circ-K\sqrt{c} \] Given: \[ \lambda^\circ_{Na^+}=50\; S\,cm^2\,mol^{-1} \] \[ \lambda^\circ_{Cs^+}=77\; S\,cm^2\,mol^{-1} \]

• A. Fig 1
• B. Fig 2
• C. Fig 3
• D. Fig 4

Hint:

For strong electrolytes: \[ \boxed{\Lambda_m=\Lambda_m^\circ-K\sqrt{c}} \] Hence:

• Plot of \(\Lambda_m\) vs \(\sqrt{c}\) is a straight line.

• Intercept at \(\sqrt{c}=0\) gives \(\Lambda_m^\circ\).

• Higher ionic mobility \(\Rightarrow\) higher \(\Lambda_m^\circ\).

Answer 1

The Correct Option is B

Concept: For strong electrolytes, \[ \Lambda_m=\Lambda_m^\circ-K\sqrt{c} \] This equation represents a straight line with: \[ \text{slope}=-K \] and y-intercept \[ \Lambda_m^\circ. \] Hence, \(\Lambda_m\) decreases linearly with \(\sqrt{c}\).

Step 1: Compare \(\Lambda_m^\circ\) values. \[ \Lambda_m^\circ(NaCl) = \lambda^\circ_{Na^+} + \lambda^\circ_{Cl^-} \] \[ \Lambda_m^\circ(CsCl) = \lambda^\circ_{Cs^+} + \lambda^\circ_{Cl^-} \] Since \[ \lambda^\circ_{Cs^+}=77 \gt \lambda^\circ_{Na^+}=50, \] we have \[ \Lambda_m^\circ(CsCl) \gt \Lambda_m^\circ(NaCl). \] Therefore, the \(CsCl\) line must lie above the \(NaCl\) line.

Step 2: Determine the nature of the graph. For strong electrolytes: \[ \Lambda_m \downarrow \text{ as } \sqrt{c} \uparrow \] Therefore the graph must have a negative slope.

Step 3: Select the correct figure. The correct graph should show:

• Straight lines with negative slope

• \(CsCl\) above \(NaCl\)

• Nearly parallel lines
This corresponds to

Option (B).

Final Answer: \[ \boxed{\text{Option (B)}} \]