Question

Using the freezing point depression method, how many grams of glucose is needed to make a 3500mL solution which contains 6g MgSO4 and 7g sodium chloride isotonic with blood?

Glucose ΔT^1% = 0.10

Sodium Chloride ΔT^1% = 0.576

MgSO₄ ΔT^1% = 0.1

Answer 1

An isotonic solution with blood menas the solution should have same osmotic pressure as that of blood. Now in pharmatutical calculations, the easier way to make a solution isotonic with blood is to make a freezing point depression of 0.52^◦C

i.e here we have to take the freezing point of blood =  0^◦C

And the freezing point of the solution = freezing point depresstion - freezing point of bloog

= 0.52^◦C - 0^◦C

= -0.52^◦C

It is calculated in the following way-

i) We alreadyd have the freezing point depression for MgSO4 present = 0.1

That means freezing point of  MgSO4 = -0.1^◦C

Similarly freezing point depression for Sodium Chloride present = 0.576^◦C

That means freezing point of  Sodium Chloride = -0.576^◦C

So the sum of freezing point of these two = -0.1^◦C + (-0.576^◦C) =  -0.676^◦C

Thus the required value of freezing point of  glucose to make the total freezing point of solution (-0.52^◦C) is-

= -0.52^◦C - (-0.676^◦C)

= 0.156‬^◦C

Or the required freezing point depression (ΔT) for the to be added glucose solution = 0.156‬^◦C

Now the formula to calculate the amount of glucose required is-

W = (0.52 - a) / b

where

W = weight of the substannce required to be added pre 100 mL of solution

a = required freezing point depression to be adjusted = 0.156‬^◦C

b = freezing point depression of the glucose solution = 0.10‬^◦C

So putting these values-

W = (0.52 - a) / b

= (0.52 - 0.156) / 0.10

= 3.64‬

or

W = 3.64‬ g/100 mL

i.e 3.64‬ g mass of glucose is required to be added per 100 mL volume of solution

Given required volume of solution = 1300 mL

So

required mass of glucose to be added = 1300 mL * 3.64‬ g/100 mL

= 47.32‬ g