In a solution containing 0.50 m sucrose in water, where Kb = 0.512°C·m^-1 and i = 1, what is the boiling point elevation ΔTb?

In a solution containing 0.50 m sucrose in water, where Kb = 0.512°C·m^-1 and i = 1, what is the boiling point elevation ΔTb?

• A. 0.128°C
• B. 1.024°C
• C. 0.256°C
• D. 0.512°C

Answer: (C)

The concept being tested is boiling point elevation, which follows ΔTb = i × Kb × m. This tells us how much the solvent’s boiling point increases when a solute is dissolved. In this case, sucrose is a non-electrolyte, so it does not dissociate in solution and the van’t Hoff factor i is 1. The given m is 0.50 molal (molality), meaning 0.50 moles of solute per kilogram of solvent. The ebullioscopic constant for water is Kb = 0.512 °C·m^-1. Plugging in the values: ΔTb = 1 × 0.512 × 0.50 = 0.256 °C. So the boiling point is raised by 0.256 °C (the final boiling point would be 100.256 °C if needed).

The concept being tested is boiling point elevation, which follows ΔTb = i × Kb × m. This tells us how much the solvent’s boiling point increases when a solute is dissolved.

In this case, sucrose is a non-electrolyte, so it does not dissociate in solution and the van’t Hoff factor i is 1. The given m is 0.50 molal (molality), meaning 0.50 moles of solute per kilogram of solvent. The ebullioscopic constant for water is Kb = 0.512 °C·m^-1.

Plugging in the values: ΔTb = 1 × 0.512 × 0.50 = 0.256 °C.

So the boiling point is raised by 0.256 °C (the final boiling point would be 100.256 °C if needed).