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Calculate the vapor pressure of a solution prepared by dissolving 0.500 mol of a non-volatile solute in 275 g of hexane ( 86.18 g/mol) at 49.6°C. P°hexane = 400.0 torr at 49.6°C.

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Related questions

1. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the vapor pressures of pentane and hexane are 511 and 150 torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 36 mL hexane (density, 0.66 g/mL). What is the vapor pressure of the resulting solution? Ptotal = ___ torr 2. Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25oC the vapor pressures of pentane and hexane are 511 and 150 torr, respectively. A solution is prepared by mixing 25 mL pentane (density, 0.63 g/mL) with 36 mL hexane (density, 0.66 g/mL). What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution? X V pen = ___

Calculate the vapor pressure (in torr) at 293 K in a solutionprepared by dissolving 43.5 g of the non-volatile non-electrolyeglucose in 159 g of water. The vapor pressure of water at 293 K is17.54 torr.

Vapour Pressure of Solutions of Non-Volatile or Volatile Solutes

31.686 g of a non-volatile solute is dissolved in 460.0 g of water.
The solute does not react with water nor dissociate in solution.
Assume that the resulting solution displays ideal Raoult's law behaviour.
At 10°C the vapour pressure of the solution is 9.055 torr.
The vapour pressure of pure water at 10°C is 9.209 torr.
Calculate the molar mass of the solute (g/mol).

See example 17.1 on pp865-6 of Zumdahl ''Chemical Principles'' 7th ed.


Now suppose, instead, that 31.686 g of a volatile solute is dissolved in 460.0 g of water.

This solute also does not react with water nor dissociate in solution.
The pure solute displays, at 10°C, a vapour pressure of 0.921 torr.
Again, assume an ideal solution.
If, at 10°C the vapour pressure of this solution is also 9.055 torr.
Calculate the molar mass of this volatile solute.

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