Question

18 gm glucose \(C_6H_{12}O_6\) is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol\(^{-1}\).

Hint:

Boiling point elevation depends on the molality of the solution and the ebullioscopic constant.

Answer 1

Step 1: Use the formula for boiling point elevation.
The boiling point elevation is given by: \[ \Delta T_b = K_b \times m \] where: - \( \Delta T_b \) is the change in boiling point, - \( K_b \) is the ebullioscopic constant for water (0.52 K kg mol\(^{-1}\)), - \( m \) is the molality of the solution.

Step 2: Calculate the molality.
Molality \( m \) is given by: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] The moles of glucose is: \[ n_{\text{glucose}} = \frac{18}{180} = 0.1 \, \text{mol} \] Since the mass of water is 1 kg, the molality is: \[ m = \frac{0.1}{1} = 0.1 \, \text{mol/kg} \]

Step 3: Calculate the change in boiling point.
\[ \Delta T_b = 0.52 \times 0.1 = 0.052 \, \text{K} \]

Step 4: Calculate the boiling point.
The boiling point of water is 100°C at 1 atm. The boiling point at 1.013 bar is: \[ T_b = 100 + \Delta T_b = 100 + 0.052 = 100.052°C \]

Step 5: Conclusion.
Thus, the boiling point of the solution at 1.013 bar is 100.052°C.