Question

180 g of glucose, \(C_6H_{12}O_6\), is dissolved in 1 kg of water in a vessel. The temperature at which water boils at 1.013 bar is ______ (given, \(K_b\) for water is 0.52 K kg mol\(^{-1}\)). Boiling point for pure water is 373.15 K

• A. 373.15 K
• B. 373.0 K
• C. 373.202 K
• D. 373.67 K

Hint:

When calculating colligative properties like boiling point elevation or freezing point depression, remember that the effect depends on the number of solute particles, not their identity.

Answer 1

The Correct Option is D

To calculate the boiling point elevation, we use the formula: \[ \Delta T_b = K_b \cdot m \] Where: - \(\Delta T_b\) is the boiling point elevation, - \(K_b\) is the ebullioscopic constant, - \(m\) is the molality of the solution. We are given: - \(w_2 = 180 \, \text{g}\) (mass of glucose), - \(M_2 = 180 \, \text{g/mol}\) (molar mass of glucose), - \(w_1 = 1 \, \text{kg}\) (mass of water), - \(K_b = 0.52 \, \text{K} \, \text{kg/mol}\) (for water). Now, calculate the molality (\(m\)): \[ m = \frac{\text{mol of solute}}{\text{mass of solvent in kg}} = \frac{w_2 / M_2}{w_1} \] \[ m = \frac{180 / 180}{1} = 1 \, \text{mol/kg} \] Now, calculate the boiling point elevation: \[ \Delta T_b = 0.52 \times 1 = 0.52 \, \text{K} \] Finally, the boiling point of the solution will be: \[ T_b = T_0 + \Delta T_b = 373.15 + 0.52 = 373.67 \, \text{K} \] Thus, the answer is \(373.67 \, \text{K}\).