Question

A pharmacist–herbalist mixed 100 g lots o St. John’s wort containing the ollowing percentages o the active component hypericin: 0.3%, 0.7%, and 0.25%. Calculate the percent strength o hypericin in the mixture.

Answer 1

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 =  $\frac{0.3}{100}$ × 100

Hypericin contribution lot 1 =  0.3 g

and

Hypericin contribution lot 2 = $\frac{0.3}{100}$ × 100

Hypericin contribution lot 2 = 0.7 g

and

Hypericin contribution lot 3 = $\frac{0.25}{100}$ × 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 = $\frac{1.25}{300}$ × 100  

strength of hypericin in mixture = 0.42 %