Identify the correct relation between the molar mass of solute and Ebullioscopic constant.
Answer & explanation
Correct answer: option 3
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. \(M_2 = \frac{1000 \times w_2 \times K_b}{\Delta T_b \times w_1}\)