Question

If an azithromycin formulation contains 500mg of azithromycin and sterile water is added to make 5ml of the formulation, how much must be removed for a dosage of 80mg

Answer 1

0.606 g L / mol is required to remove for a dosage of 80 mg.

Mass of azithromycin = 500 mg = 0.5 g

The volume of sterile water = 5 ml

number of moles = n

n = weight of azithromycin / molar mass of azithromycin

n = 0.5 / 749  ------  since the molar mass of azithromycin is 749 g / mol

n = 6.6 * 10⁻⁴

concentration = n / V

                      = 6.6 * 10⁻⁴ / 50 * 10⁻⁴

                      = 0.132 mol / L

Now, we know that

liquid suspension = a = dose / concentration

a = 80 * 10⁻³ / 0.132

 = 606.06 * 10⁻³

 = 0.606 g L / mol

0.606 g L / mol is required to remove for a dosage of 80 mg.

For more information on Concentration click on the link below:

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