Question

For a salt XY, which is a strong electrolyte, the plot of \(\Lambda_m\) versus \(\sqrt{c}\) has a slope of \(-90.0\text{ S cm}^2\text{ mol}^{-3/2}\text{ L}^{1/2}\) at \(298\text{ K}\). At \(0.01\text{ M}\) concentration of XY, the value of \(\Lambda_m\) is \(145.0\text{ S cm}^2\text{ mol}^{-1}\). The limiting molar conductivity of \(\text{Y}^-\) ion (\(\lambda^\circ_{\text{Y}^-}\), in \(\text{S cm}^2\text{ mol}^{-1}\)) at \(298\text{ K}\) will be (Given: \(\lambda^\circ_{\text{X}^+} = 74.0\text{ S cm}^2\text{ mol}^{-1}\))

• A. \(76.0\)
• B. \(80.0\)
• C. \(100.0\)
• D. \(90.0\)

Hint:

Be extra precise when calculating the square root of concentration decimals. For example, $\sqrt{0.01} = 0.1$, not $0.01$. Once you successfully solve for the total limiting conductance $\Lambda_m^\circ$, simply subtract the value of the known ion to get the unknown ion instantly.

Answer 1

The Correct Option is B

Step 1: Recognise the working equation.
For a strong electrolyte the Debye-Huckel-Onsager law gives \[ \Lambda_m = \Lambda_m^\circ - A\sqrt{c} \] where the slope of \(\Lambda_m\) versus \(\sqrt{c}\) equals \(-A\).
Step 2: Read off the slope.
The given slope is \(-90.0\), so \(A = 90.0\).
Step 3: Compute \(\sqrt{c}\).
At \(c = 0.01\) M, \(\sqrt{0.01} = 0.1\).
Step 4: Solve for the limiting molar conductivity of the salt.
Substituting the measured \(\Lambda_m = 145.0\): \[ 145.0 = \Lambda_m^\circ - 90.0 \times 0.1 = \Lambda_m^\circ - 9.0 \] so \(\Lambda_m^\circ = 145.0 + 9.0 = 154.0\,S\,cm^2\,mol^{-1}\).
Step 5: Apply Kohlrausch's law for the 1:1 salt XY.
\[ \Lambda_m^\circ(XY) = \lambda^\circ_{X^+} + \lambda^\circ_{Y^-} \]
Step 6: Isolate the anion conductivity.
With \(\lambda^\circ_{X^+} = 74.0\): \[ \lambda^\circ_{Y^-} = 154.0 - 74.0 = 80.0\,S\,cm^2\,mol^{-1} \]
\[ \boxed{80.0} \]