Answer:
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
We have that for the Question "How many mg of mupirocin powder are required? "
It can be said that
From the question we are told
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
Weight of standard package = 22g
mupirocin = 2%
Therefore,
Weight of mupirocin = [tex]\frac{2}{100}*22[/tex]
[tex]=0.44g[/tex]
Let amount of mupirocin to be added to prepare 3% w/w mupirocin ointment = X
Therefore,
Total weight of ointment = [tex](22 + X)g[/tex]
amount of mupirocin = [tex](0.44+X)g[/tex]
% of mupirocin in ointment = [tex]3\%[/tex]
therefore,
[tex]\frac{3}{100} = \frac{0.44+X}{22+X}\\\\66+3X = 44+100X\\\\X = 0.2268g[/tex]
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