For strong electrolyte $$\lambda_m$$ increases slowly with dilution and can be represented by the equation
$$\Lambda_m = \Lambda_m^\circ - A c^{1/2}$$
Molar conductivity values of the solutions of strong electrolyte AB at 18°C are given below:

The value of constant A based on the above data [in S $$cm^{2}mol^{-1}/(mol/L)^{1/2}$$]unit is_______.

We need to determine the value of the constant $$A$$ from the given molar conductivity data using the Debye-Hückel-Onsager equation.

1. Formula and Data Points:

The relationship is given by the equation:

$$\Lambda_m = \Lambda_m^\circ - A c^{1/2}$$

Let's choose two data points from the table to eliminate the limiting molar conductivity ($$\Lambda_m^\circ$$):

• Point 1: At $$c_1 = 0.04\text{ mol L}^{-1}$$, $$\Lambda_{m1} = 96.1\text{ S cm}^2\text{ mol}^{-1}$$
  $$\implies c_1^{1/2} = \sqrt{0.04} = 0.2$$
• Point 2: At $$c_2 = 0.25\text{ mol L}^{-1}$$, $$\Lambda_{m2} = 94.9\text{ S cm}^2\text{ mol}^{-1}$$
  $$\implies c_2^{1/2} = \sqrt{0.25} = 0.5$$

2. Calculation of Constant A:

Subtracting the two equations to solve for slope ($$A$$):

$$\Lambda_{m1} - \Lambda_{m2} = A\left(c_2^{1/2} - c_1^{1/2}\right)$$

$$96.1 - 94.9 = A(0.5 - 0.2)$$

$$1.2 = A(0.3)$$

$$A = \frac{1.2}{0.3} = 4$$

Conclusion:

The value of the constant $$A$$ is exactly 4.

Answer: 4