Question

a. Assuming that glucose and water form an ideal solution, what is the partial pressure of water at 20o C of a solution of 1.00 g of glucose (mol wt 180 g/mol) in 100 g of water? The vapor pressure of pure water is 2338 Pa at 20o C.

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

c. What would be the osmotic pressure, in Pa, of a solution containing both 1.00 g of glucose and 1.00 g of sucrose (mol wt 342) in 100 g of water at 20o C?

Answer 1

a. Assuming that glucose and water form an ideal solution, what is the partial pressure of water at 20o C of a solution of 1.00 g of glucose (mol wt 180 g/mol) in 100 g of water? The vapor pressure of pure water is 2338 Pa at 20o C.

dP = xsolute * Psolvent

xsolute = mol of solute / total mol

mol of solute = mass/MW = 1/180 = 0.00555 mol

mol of water = 100 g = mass/MW = 100/18 = 5.555 mol

total mol = 0.00555+5.555 = 5.56055 mol

xsolute = 0.00555 /5.56055 = 0.000998

dP = xsolute * Psolvent

dP =0.000998*2338 = 2.3333 Pa

Pmix = Psolvent - dP = 2338 - 2.3333

Pmix = 2335.6667 Pa

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

Psomitc = M*R*T

M = mol of solute / volume

m = 100 g --> 100 mL = 0.1 L

M = 0.00555 /0.1 = 0.0555

T = 20°C, 293 K

Posm = (0.0555 )(0.082)(293 )= 1.333443 atm

change to Pa

Posm = 1.333443 *101325 = 135111.11 Pa

c. What would be the osmotic pressure, in Pa, of a solution containing both 1.00 g of glucose and 1.00 g of sucrose (mol wt 342) in 100 g of water at 20o C?

total mol --> mol of glucose + mol of sucrose

mol of glucose = 1/180 = 0.00555

mol of sucrose = 1/342 = 0.00292

Psomitc = M*R*T

Psomitc = (0.00555+0.00292)/0.1 * 0.082*293

Posm = 2.035 atm

Posm = 2.035*101325 = 206196.375 Pa