You need to prepare at 2.0 ml sample of a diluted drug for injection. the total amount of the drug to be injected in this 2.0 ml injection is 0.0210 mg of drug/kg-body mass. assume the average body mass of 70 kg. the label tells you that the volume of solution in the drug bottle is 30.0 ml. the total mass of drug in the bottle is 294 mg and the rest is saline (a salt solution). in addition to this bottle of concentrated drug, you have an unlimited supply of pure, sterile saline to serve as a diluent. a. what is the concentration (mg/ml) of drug in the bottle? b. what volume of concentrated drug (in ml) and what volume of saline (in ml) will you combine to make a 2.0 ml final volume of the drug solution at the necessary concentration? c. the molecular weight of the drug is 15,000 g/mol. what is the molarity of the injection?

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

the correct answer is:

a. formation of precipitation