Question

Consider a solution of 1.00 g of glucose in 100 g of water at 20˚C (where the vapor pressure of pure water is 17.54 torr). For this problem we will assume the solution behaves ideally.

(a) What is the equilibrium vapor pressure of the solution?

(b) What is the osmotic pressure of the solution versus pure water (report it in torr)?

(c) What is the activity of water as a solvent in the solution?

(d) Now suppose 1.00 g of sucrose is added to the solution. Repeat part (b) for this new solution.

e) Consider two solutions: Soln. I: 1.0M sucrose in water Soln. II: 0.5M sucrose in water. The two solutions are in separate (open) beakers. The two beakers are now placed next to each other in a sealed jar at constant temperature. Will anything happen? If so, describe what will happen and why. If not, why is the system already in equilibrium?

Answer 1

1 g glucose in 100 g water

(a) moles glucose = 1 g/180.156 g/mol = 0.0055 mol

moles water = 100 g/18.02 g/mol = 5.55 mol

Total moles = 5.5555 mol

mole fraction of water = 5.55/5.5555 = 0.9991

According to Raoults law,

vapor pressure of solution = mole fraction x vapor pressure of pure solvent

                                          = 0.9991 x 17.54 torr = 17.52 torr

(b) molarity of solution = 0.0055 mol/0.1 L = 0.055 M

osmotic pressure = MRT

R = gas constant

So,

osmotic pressure = 0.055 mol/L x 0.0821 L.atm/K.mol x (25 + 273) K

                            = 1.346 atm x 760 torr = 1022.96 torr

(c) activity of water in solution = 1

(d) moles of sucrose = 1 g/342.3 g/mol = 0.003 mol

molarity of solution = 0.003 mol/0.1 L = 0.03 M

osmotic pressure = 0.03 x 0.0821 x 298 x 760 = 557.84 torr

(e) If the two solutions are kept together, the solvent from the less dilute solution (0.5 M) will move towards high concentrated solution (1.0 M) until the concentrations of both the solutions are equal.