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 sucrose, C,H0, and calcium chloride. STRATEGY 1. Determine the colligative molality of each solution. 2. Convert each colligative molality to a mole fraction. 3. Apply Raoult's law. Step 1: Sucrose is a nonelectrolyte whereas CaCl, has 3 ions per formula unit. sucrose: 2.40 m • CaCl,: 3(2.40 m) = 7.20 m %3D Step 2: Molality is the number of moles of solute per kilogram of solvent. 1000 g 1 kg H,O x 1 mol 18.0148 g 55.5099 mol H,O %3! 1 kg 2.40 mol Xsucrose = = 0.0414 (2.40 + 55.5099) mol 7.20 mol = 0.115 %3D %3D (7.20 +55.5099) mol Step 3: Calculate the vapor-pressure lowering at 20 °C.

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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 sucrose, C,2H,,0, and calcium chloride.
STRATEGY
1. Determine the colligative molality of each solution.
2. Convert each colligative molality to a mole fraction.
3. Apply Raoult's law.
Step 1: Sucrose is a nonelectrolyte whereas CaCl, has 3 ions per formula unit.
sucrose: 2.40 m
CaCl,: 3(2.40 m) = 7.20 m
Step 2: Molality is the number of moles of solute per kilogram of solvent.
1000 g
1 mol
1 kg H,0 x
1 kg
= 55.5099 mol H,0
18.0148 g
2.40 mol
Xsucrose =
= 0.0414
(2.40 + 55.5099) mol
7.20 mol
XCaCl, =
= 0.115
(7.20 + 55.5099) mol
Step 3: Calculate the vapor-pressure lowering at 20 'C.
torr
sucrose, AP =
torr
calcium chloride, AP =
Question Source: MRG General Chemistry
MacBook Ai
000
F9
F10
F11
D00
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F7
F8
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6
Transcribed Image Text: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 sucrose, C,2H,,0, and calcium chloride. STRATEGY 1. Determine the colligative molality of each solution. 2. Convert each colligative molality to a mole fraction. 3. Apply Raoult's law. Step 1: Sucrose is a nonelectrolyte whereas CaCl, has 3 ions per formula unit. sucrose: 2.40 m CaCl,: 3(2.40 m) = 7.20 m Step 2: Molality is the number of moles of solute per kilogram of solvent. 1000 g 1 mol 1 kg H,0 x 1 kg = 55.5099 mol H,0 18.0148 g 2.40 mol Xsucrose = = 0.0414 (2.40 + 55.5099) mol 7.20 mol XCaCl, = = 0.115 (7.20 + 55.5099) mol Step 3: Calculate the vapor-pressure lowering at 20 'C. torr sucrose, AP = torr calcium chloride, AP = Question Source: MRG General Chemistry MacBook Ai 000 F9 F10 F11 D00 F4 F6 F7 F8 %24 & 5 7 8. 6
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