Question

(1) The vapor pressure of pure water is 23.8 mmHg at 25.0 °C. What is the vapor pressure of 2.50 molal C6H12O6 (2) What is the vapor pressure of an aqueous solution that has a solute mole fraction of 0.1000? The vapor pressure of water is 25.756 mmHg at 25 °C. (3) The vapor pressure of an aqueous solution is found to be 24.90 mmHg at 25 °C. What is the mole fraction of solute in this solution? The vapor pressure of water is 25.756 mm Hg at 25 °C. (4) At 29.6 °C, pure water has a vapor pressure of 31.1 torr. A solution is prepared by adding 86.8 g of "Y", a nonvolatile non-electrolyte to 350. g of water. The vapor pressure of the resulting solution is 28.6 torr. Calculate the molar mass of Y.

Answer 1

Answer:

First convert the molality to a mole fraction

⇒2.50m C

6

H

12

O

6

=2.50 mol/1kg of H

2

O

⇒1000g/18.015=55.51 mol

⇒55.51+2.50=58.01 mol

Mole fraction of H

2

O=

58.01

55.81

=0.9569

Raoult's Law:

P

solution

=(Ysolvent)(P

o

solvent)

=(0.9569)(23.8)=22.8mmHg