As a part of a clinical study, a pharmacist is asked to prepare a modification of a standard 22g package of a 2% mupirocin ointment by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment. How many mg of mupirocin powder are required?

226.8 mg of mupirocin powder are required

Explanation:

Given that;

weight of standard pack = 22 g

mupirocin by weight = 2%

so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g

so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment

mg of mupirocin powder are required = ?, lets rep this with x

Total weight of ointment = 22 + x g

Amount of mupirocin  = 0.44 + x g

percentage of mupirocin  in ointment is 3?

so

3/100 = 0.44 + x g / 22 + x g

3( 22 + x g ) = 100( 0.44 + x g )

66 + 3x g = 44 + 100x g

66 - 44 = 100x g - 3x g

97 x g = 22

x g = 22 / 97

x g = 0.2268 g

we know that; 1 gram = 1000 Milligram

so 0.2268 g = x mg

x mg = 0.2268 × 1000

x mg  = 226.8 mg

Therefore, 226.8 mg of mupirocin powder are required

b

explanation:

mitosis

it's c. elements are placed in rows based on how may valencies electrons have.