What osmotic pressure is exerted by 1.50 M glucose solution at 22.0 degrees C?
a. 5.32 x 10 1 kPa
b. 2.45 x 10 3 kPa
c. 3.68 x 10 3 kPa
d. 2.74 x 10 2 kPa

Hi again!!

Osmotic Pressure = M.R.T  (M = Molarity, R = 0.08206 L x atm/K x mol,
T = temp in K)

In this problem the molarity is 1.5 and the temperature in Kelvin
degrees is 295ēK (=273+22).

Osmotic Pressure = M.R.T = 1.5x(.08206)x(273+22) = 36.31 atm 
1 atm is equivalent to 101.325 kPa; then 
36.31 atm are equivalent to 36.31 x 101.325 kPa = 3679 kPa = 3.68 x
10^3 kPa.

The correct answer is c.