Sankalp NEET Full Test-3 Question-61 Solution

Question: 61. If the molar conductivity \left(\Lambda_{\mathrm{m}}\right) of a 0.050 \mathrm{~mol} \mathrm{~L}^{-1} solution of a monobasic weak acid is 90 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}, its extent (degree) of dissociation will be [Assume \Lambda_{+}^{\circ}=349.6 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} and \Lambda_{-}^{\circ}=50.4 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} ]

(1) 0.115

(2) 0.125

(3) 0.225

(4) 0.215

Answer: Option (3)

Explanation:

For a weak electrolyte (here, a monobasic weak acid), the degree of dissociation \alpha is related to molar conductivity by:

\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}

First, calculate the limiting molar conductivity \Lambda_{m}^{\circ} of the acid using the limiting ionic conductivities of its ions:

\Lambda_{m}^{\circ} = \Lambda_{+}^{\circ} + \Lambda_{-}^{\circ}

Given:

\Lambda_{+}^{\circ} = 349.6 \Lambda_{-}^{\circ} = 50.4

So,

\Lambda_{m}^{\circ} = 349.6 + 50.4 = 400.0

Now use the given molar conductivity of the solution:

\Lambda_{m} = 90

Substitute in the formula for \alpha:

\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{90}{400}

Calculate the value:

\alpha = \frac{90}{400} = 0.225

Therefore, the degree of dissociation of the weak acid is 0.225,

which corresponds to Option (3).