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vagabundo [1.1K]
3 years ago
11

The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 at

m at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? g
Chemistry
2 answers:
malfutka [58]3 years ago
8 0

Answer:

0.291

Explanation:

Given data

  • Vapor pressure of the pure solvent (Psolvent): 1.41 atm

1070 torr × (1 atm/ 760 torr) = 1.41 atm

  • Vapor pressure of the solvent above the solution (Psolution): 1.00 atm

According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present.

Psolution = Psolvent × Χsolvent

Χsolvent = Psolution/Psolvent

Χsolvent = 1.00 atm/1.41 atm

Χsolvent = 0.709

The sum of the mole fraction of the solvent (water) and the solute (ethylene glycol) is 1.

Χsolvent + Χsolute = 1

Χsolute = 1 - Χsolvent = 1 - 0.709

Χsolute = 0.291

Kitty [74]3 years ago
7 0

Answer:

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