Respuesta :
Answer:
strength of hypericin in mixture = 0.42 %
Explanation:
given data
each lot = 100 g
active component hypericin = 0.3%, 0.7%, and 0.25%
solution
we get here percent strength o hypericin in the mixture that is
Hypericin contribution lot 1 = [tex]\frac{0.3}{100}[/tex] × 100
Hypericin contribution lot 1 = 0.3 g
and
Hypericin contribution lot 2 = [tex]\frac{0.3}{100}[/tex] × 100
Hypericin contribution lot 2 = 0.7 g
and
Hypericin contribution lot 3 = [tex]\frac{0.25}{100}[/tex] × 100
Hypericin contribution lot 3 = 0.25 g
so
total 300 g mixture of hypericin contain = 0.3 g + 0.7 g + 0.25 g
total 300 g mixture of hypericin = 1.25 g
so here percent strength o hypericin in mixture is
strength of hypericin in mixture = [tex]\frac{1.25}{300}[/tex] × 100
strength of hypericin in mixture = 0.42 %
The percent strength of hypericin in the mixture will be "0.42%".
According to the question,
From lot 1, Hypericin contribution will be:
= [tex]100\times \frac{0.3}{100}[/tex]
= [tex]0.3 \ g[/tex]
From lot 2, Hypericin contribution will be:
= [tex]100\times \frac{0.7}{100}[/tex]
= [tex]0.7 \ g[/tex]
From lot 3, Hypericin contribution will be:
= [tex]100\times \frac{0.25}{100}[/tex]
= [tex]0.25 \ g[/tex]
For 300 g mixture,
The amount of hypericin will be:
= [tex]0.3+0.7+0.25[/tex]
= [tex]1.25 \ g[/tex]
hence,
The percentage strength will be:
= [tex]\frac{1.25}{300}\times 100[/tex]
= [tex]0.42[/tex] (%)
Thus the above approach is right.
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