Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: boostar on May 02, 2008, 01:39:21 PM

Title: Isotonicity and Freezing Point Depression
Post by: boostar on May 02, 2008, 01:39:21 PM
Hi all, I having particular trouble with a problem that revolves around isotonicity and freezing point depression. Any help would be appreciated.

6) Zinc & Adrenaline Eye Drops
Zinc Sulphate                          0.25g
Adrenaline                              0.02g
Boric acid                                as required
Glycerol                                  1 mL
Purified water                         to 100mL                   

Use the following information to make 25.0mL of these eye drops for a patient and calculate the required amount of boric acid required to make the solution isotonic.
You already have an adrenaline 1:1000 solution that is isotonic (assume this is used to make up the eye drops).
FD1% Zinc Sulphate   = 0.086 °C
FD1% Boric acid  = 0.288 °C
FD1% Glycerol = 0.203 °C
Express your answer in milligrams to 3 significant figures.

Worked Answer:

Zinc Sulphate                          0.25 %(w/v)
Adrenaline                              0.02g
Boric acid                                as required
Glycerol                                  1 %(v/v)
Purified water                         to 100mL       

So, we need it to be isotonic so Freezing point must be -0.52oC

0.52- (0.086/4) - 0.203 = 0.2955

than divide by FD1% Boric acid

0.2955/0.288 =  1.026%(w/v) of Boric Acid

The questions asks for Boric acid in 20mL
Adrenaline is already isotonic, find it's volume in 20mL -> 0.02/5 * 1000 = 5mL
Therefore, 25mL-5mL must be made isotonic (20mL)

Therefore 1.026/5 = 0.2052g * 1000 =205mg of Boric Acid must be added.

Apparently however, this is wrong! Could someone please point out my error?

Many thanks!