The formula for boiling point elevation is given by: \[ \Delta T_b = K_b \cdot m \cdot i \] where:
\( \Delta T_b \) is the boiling point elevation,
\( K_b \) is the ebullioscopic constant,
\( m \) is the molality of the solution,
\( i \) is the van't Hoff factor (number of particles produced per formula unit).
Given that \( \Delta T_b = 0.3^\circ \text{C} \), \( K_b = 0.52 \, \text{K kg mol}^{-1} \), and the solute 'X' dimerizes in water to the extent of 80%, the van't Hoff factor \( i = 1 + 0.8 = 1.8 \). Now, molality \( m \) is given by: \[ m = \frac{\text{mol of solute}}{\text{mass of solvent in kg}} \] The mass of the solvent is 100g, or 0.1kg. The number of moles of 'X' is: \[ \text{moles of X} = \frac{\text{mass of X}}{\text{molar mass of X}} = \frac{2.5}{M} \] where \( M \) is the molar mass of 'X'. Substitute into the boiling point elevation formula: \[ 0.3 = 0.52 \cdot \left( \frac{2.5}{M \cdot 0.1} \right) \cdot 1.8 \] Simplifying: \[ 0.3 = \frac{0.936 \cdot 2.5}{M} \] \[ M = \frac{0.936 \cdot 2.5}{0.3} = 26 \]
Thus, the molar mass of 'X' is 26 g/mol.
The correct option is (A) : 26
Given that the solute ‘X’ dimerises to 80%, the van't Hoff factor ($i$) will be calculated based on the dimerisation. If 80% of the solute dimerises, then the effective number of particles decreases. The formula for the boiling point elevation is given by: \[ \Delta T_b = i \cdot K_b \cdot m \] Where: \begin{itemize} \item \(\Delta T_b\) = Boiling point elevation = 0.3°C \item \(K_b\) = ebullioscopic constant = 0.52 K kg mol$^{-1}$ \item \(m\) = molality = \(\frac{mol \text{ of solute}}{kg \text{ of solvent}}\) \end{itemize} The molality is given by: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{\frac{2.5}{M}}{0.1} \] Where \(M\) is the molar mass of the solute. Substituting into the formula for the boiling point elevation: \[ 0.3 = i \cdot 0.52 \cdot \frac{\frac{2.5}{M}}{0.1} \] Since the solute dimerises by 80%, the van't Hoff factor \(i = 1 + 0.8 = 1.8\). Thus: \[ 0.3 = 1.8 \cdot 0.52 \cdot \frac{\frac{2.5}{M}}{0.1} \] Solving for \(M\), we get: \[ M = \frac{1.8 \cdot 0.52 \cdot 2.5}{0.3 \cdot 0.1} = 26 \] Thus, the molar mass of ‘X’ is 26 g/mol.
14g of cyclopropane burnt completely in excess oxygen. The number of moles of water formed is: