Question

3. (a) Assuming that glucose and water form an ideal solution, what is the equilibrium vapor pressure (in torr) at 20°C of a solution of 1.00 g of glucose (molar mass 180 g/mol) in 100 g of water? The vapor pressure of pure water is 17.54 torr at 20°C.

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

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

(d) What would be the osmotic pressure (in torr) versus pure water of a solution containing both 1.00 g of glucose and 1.00 g of sucrose (molar mass 342 g/mol) in 100 g of water at 20°C?

Answer 1

1) No. Of moles of glucose n1 = mass/molar mass = 1g/180 g/mol = 0.0056

No. Of moles of water n2 = mass/molar mass = 100 g/18.02 g/mol = 5.55 mol

Mole fraction of glucose x1 = n1/n1+n2 = 0.0056/5.56 = 0.001

Mole fraction of water x2 = n2/n1+n2 = 5.55 /5.56 = 0.998

Vapor Pressure of solution = Psol = x2 P^o2 = 0.998x 17.54 torr = 17.51 torr

Relative lowering of pressure = 17.54 - 17.51 torr = 0.03 torr

2) molarity of solution = no. Of moles of solute/vol in lts

Density of water = 1 g/ml

So 100 g of water = 100 ml of water = 100/1000 = 0.1 lt

Molarity of solution = 0.0056 mol/0.1 lt. = 0.056 M

(osmotic pressure) = MRT

T= 20+273 = 293 K

R = 62.364 lt torr/mol K

= 0.056 M x 62.364 lt torr/ mol K x 293 K = 1023.27 torr

3) activity of water as a solvent a _w = P/P0 = 17.51/17.54 = 0.998

4) no. Of moles of glucose (calculated previously) = 0.0056

No. Of moles of sucrose = 1 g/342 g/mol = 0.0029

Vol of solvent = 100 g = 100 ml = 0.1 lt

Molarity of solution = no. Ofmoles of solute/vol in lts =

= 0.0056 + 0.0029 /0.1 lt = 0.085 mol/lt

= MRT

= 0.085 M x 293 K x 62.364 lt torr/mol K = 1552.58 torr