If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is th
Practice Questions
Q1
If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.52 °C kg/mol) (2021)
0.26 °C
0.52 °C
1.04 °C
0.78 °C
Questions & Step-by-Step Solutions
If 0.5 moles of a non-volatile solute are dissolved in 1 kg of water, what is the expected change in boiling point? (Kb for water = 0.52 °C kg/mol) (2021)
Step 1: Identify the formula for boiling point elevation, which is ΔT_b = i * K_b * m.
Step 2: Determine the values needed for the formula: i (van 't Hoff factor) for a non-volatile solute is 1, K_b for water is 0.52 °C kg/mol, and m (molality) is 0.5 moles in 1 kg of water.
Step 3: Plug the values into the formula: ΔT_b = 1 * 0.52 * 0.5.
Step 4: Calculate the result: ΔT_b = 0.26 °C.
Step 5: The expected change in boiling point is 0.26 °C.
Boiling Point Elevation – The phenomenon where the boiling point of a solvent increases when a non-volatile solute is added, calculated using the formula ΔT_b = i * Kb * m.
Colligative Properties – Properties that depend on the number of solute particles in a solution, not the identity of the solute.
