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

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**

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

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dissolved in 1.000L of water at 20.0 °C? The vapor pressure of pure
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2) Calculate the freezing point of a solution containing 0.43 of
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(On the answer sheet my teacher has 17.1torr but doesnt have
how he got that)

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STEP-BY-STEP?

The vapor pressure of pure water at 25.0 oC is 23.8 torr. What
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