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
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
How many grams of glucose must be dissolved in 300.0 ml of water at 25 degrees...
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