Question

Given that the vapor pressure of water is 17.54 Torr at 20 °C, calculate the vapor-pressure lowering of aqueous solutions that are 2.40 m in (a) sucrose, C12H22O11, and (b) calcium chloride. Assume 100% dissociation for electrolytes.

Answer 1

1. Calculate the colligative molality of each.
Sucrose: i = 1. mc = m*i = 2.4 * 1 = 2.4
CaCl2: i = 3. mc = 2.4 * 3 = 7.2

2. Calculate the moles of water.
Assume there are 1000g of water.
1000 g H2O * (1 mol / 18 g) = 55.5 moles of H2O

3. Calculate the mole fraction of each.
Sucrose: X = 2.4 mol / (2.4 + 55.5 mol) = 0.04145
CaCl2: X = 7.2 mol / (7.2 + 55.5 mol) = 0.11483

4. Calculate the vapor pressure lowering using Raoult's Law: ΔP = X*P°.
Sucrose: ΔP = (0.04145)(17.54 Torr) = 0.7270 Torr
CaCl2: ΔP = (0.11483)(17.54 Torr) = 2.0141 Torr