Question

Calculate the boiling point of a solution containing 18 g of glucose (C\(_6\)H\(12\)O\(_6\)) in 100 g of water. (K₋b = 0.52°C·kg/mol, Molar mass of glucose = 180 g/mol)

• A. 100.52°C
• B. 100.26°C
• C. 100.13°C
• D. 101.00°C

Hint:

To calculate the boiling point elevation, use the formula \(\Delta T_b = K_b \times m\), where \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution.

Answer 1

The Correct Option is A

The change in boiling point (\(\Delta T_b\)) is given by: \[ \Delta T_b = K_b \times m \] where \(m\) is the molality of the solution, and \(K_b\) is the ebullioscopic constant. First, calculate the molality \(m\): \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] Number of moles of glucose: \[ \text{moles of glucose} = \frac{\text{mass of glucose}}{\text{molar mass of glucose}} = \frac{18}{180} = 0.1 \, \text{mol} \] Mass of water in kg: \[ \text{mass of water} = 100 \, \text{g} = 0.1 \, \text{kg} \] So, the molality is: \[ m = \frac{0.1}{0.1} = 1 \, \text{mol/kg} \] Now, calculate the change in boiling point: \[ \Delta T_b = 0.52 \times 1 = 0.52^\circ \text{C} \] The boiling point of pure water is 100°C, so the boiling point of the solution is: \[ T_b = 100 + 0.52 = 100.52^\circ \text{C} \] Final answer
Answer: \(\boxed{100.52^\circ \text{C}}\)