Answer to Question #165766 in Chemistry for Phyroe

Question

Determine the vapor pressure of 20 degrees celsius of a cane sugar solution C12H22O11 containing 10 g of sugar and 100 g of water.

Answer 1

Solution:

The vapor pressure of pure water is 17.5 torr at 20.0 °C

(http://www.wiredchemist.com/chemistry/data/vapor-pressure).

Determine moles of water and sugar:

100 g water × (1 mol water / 18.0153 g water) = 5.5508 mol water

10 g sugar × (1 mol sugar / 342.3 g sugar) = 0.0292 mol sugar

Determine the mole fraction of the solvent (H₂O):

χ_solvent = mol water / (mol water + mol sugar)

χ_solvent = 5.5508 mol / (5.5508 mol + 0.0292 mol) = 0.99477

χ_solvent = 0.99477

Using Raoult's Law, determine the vapor pressure:

P_solution = χ_solvent × P°_solvent

P_solution = 0.99477 × 17.5 torr = 17.408 torr = 17.4 torr

P_solution = 17.4 torr

Answer: The vapor pressure (P_solution) of a cane sugar solution is 17.4 torr.