Question

How many grams of glucose must be dissolved in 300.0 ml of water at 25 degrees celsius to give an osmotic pressure of 9.05atm

Answer 1

Osmatic pressure = cRT

Where

c is the molar concentration of solution (mol/litre)

R = 0.08206 Latm mol^-1 K^-1

T = 298 kelvin

By substituting all these values into the above equation

Glucose molecular weight = 180 g mol^-1

Osmatic pressure = (w/Molecular weight) * RT

Now, we need to find out w.

9.05 atm = (w/180 gmol^-1 )*0.08206 L atm mol^-1 K^-1 * 298K

So, w= (9.05*180)/(0.08206*298)

W = 66.6 g dissolved in 1000 ml.

For 300 ml, the amount og glucose will be 19.98 g

So, 19.98 grams of glucose must be dissolved in 300.0 ml of water at 25 degrees celsius to give an osmotic pressure of 9.05atm