Identify the correct relation between the molar mass of solute and Ebullioscopic constant.

\(M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}\)

The correct answer is option 3. \(M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}\).

To identify the correct relation between the molar mass of the solute (\(M_2\)) and the ebullioscopic constant (\(K_b\)), let us start by analyzing the fundamental equations and concepts related to boiling point elevation.

The formula for boiling point elevation (\(\Delta T_b\)) is given by:

\(\Delta T_b = K_b \times m\) -----(1)

where:

\(\Delta T_b\) is the boiling point elevation,

\(K_b\) is the ebullioscopic constant,

\(m\) is the molality of the solution.

Molality (\(m\)) is defined as:

\(m = \frac{n_2}{w_1} \times 1000\) ------(2)

where:

\(n_2\) is the number of moles of solute,

\(w_1\) is the mass of the solvent in kilograms,

The factor \(1000\) converts the molality from \( \text{mol/kg} \) to \( \text{mol/1000 g} \).

The number of moles of solute (\(n_2\)) is:

\(n_2 = \frac{w_2}{M_2}\) ----(3)

where:

\(w_2\) is the mass of the solute,

\(M_2\) is the molar mass of the solute.

From equation (2) and (3)

\(m = \frac{\frac{w_2}{M_2}}{w_1} \times 1000 = \frac{1000 \times w_2}{M_2 \times w_1}\)

Substituting this into equation (1) weget

\(\Delta T_b = K_b \times \frac{1000 \times w_2}{M_2 \times w_1}\)

or, \(M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}\)

Based on the above derivation, the correct relation between the molar mass of the solute (\(M_2\)) and the ebullioscopic constant (\(K_b\)) is:

\(M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}\)

So, the correct option is:3. \(M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}\)