CUET · CHEMISTRY · PYQ PAPER 2025
Read the passage carefully and answer the questions The resistance ( \(R\) ) of an object is given by the expression \(R=\rho \frac{l}{A}\).
The inverse of \(R\) is called conductance and the inverse of \(\rho\) is called conductivity or specific conductance represented by \(\kappa\).
The conductivity depends upon the nature of the material (insulator, semiconductor or conductor) and the temperature of the substance.
Similarly, the conductance due to the ions present in the solution (ionic conductance) depends upon the nature of the electrolyte, size of ions, the nature of the solvent and the temperature.
Molar conductivity ( \(\Lambda_m\) ) of a solution at a given concentration is the conductance of volume \(V\) containing one mole of electrolyte kept between two electrodes of area \(A\) and unit distance apart. It is given by the expression \(\Lambda_m=\kappa V .\)
Molar conductivity increases with decrease in concentration and when concentration approaches zero, it is called limiting molar conductivity.
The decrease is linear for a strong electrolyte. However, for a weak electrolyte the change is gradual at higher concentrations and sharp near zero concentration.
Using limiting molar conductivity values of strong electrolytes we can calculate the limiting molar conductivity of weak electrolytes using Kohlrausch law of independent migration of ions.
From molar conductivity and limiting molar conductivity we can determine the degree of dissociation \(\left(\frac{\Lambda_m}{\Lambda_m^0}\right)\) and the ionization constant of a weak acid.
For a weak acid like acetic acid, the correct expression for the ionisation constant at concentration \(c\) is
- A \(K_a=\frac{c \Lambda_m^0}{\Lambda_m^2\left(\Lambda_m^0-\Lambda_m\right)}\)
- B \(K_a=\frac{c \Lambda_m^2}{\Lambda_m^0\left(\Lambda_m^2-\Lambda_m\right)}\)
- C \(K_a=\frac{c \Lambda_m^2}{\Lambda_m^0\left(\Lambda_m^0-\Lambda_m\right)}\)
- D \(K_a=\frac{c \Lambda_m^2}{\Lambda_m^0\left(\Lambda_m-\Lambda_m^2\right)}\)
Answer & Solution
Correct Answer
(C) \(K_a=\frac{c \Lambda_m^2}{\Lambda_m^0\left(\Lambda_m^0-\Lambda_m\right)}\)
Step-by-step Solution
Detailed explanation
\(\alpha = \frac{\Lambda_m}{\Lambda_m^0}\) \(K_a = \frac{c\alpha^2}{1-\alpha}\) \(K_a = \frac{c\left(\frac{\Lambda_m}{\Lambda_m^0}\right)^2}{1-\frac{\Lambda_m}{\Lambda_m^0}}\) \(K_a = \frac{c\frac{\Lambda_m^2}{(\Lambda_m^0)^2}}{\frac{\Lambda_m^0-\Lambda_m}{\Lambda_m^0}}\)…
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