Question

What percent of a drug is contained in a mixture of powder consisting of 0.5 kg, containing 0.038% of a drug, and 10 kg, containing 0.043% of a drug?

Answer 1

`0.0427~%`

Step-by-step explanation:

In this question, we have to start with the calculation of the amount of drug in each powder:

Powder A: Total mass of 0.5 Kg percentage of 0.038%

`0.5~Kg~of~powder(0.038)/(100)=0.00019~Kg~of~drug`

Powder B: Total mass of 10 Kg percentage of 0.043%

`10~Kg~of~powder(0.043)/(100)=0.0043~Kg~of~drug`

The total mass of powder would be:

`10+0.5=10.5~Kg`

The total mass of drug would be:

`0.0043+0.00019=0.00449~Kg`

Now we can calculate the percentage:

`(0.0043Kg)/(10.5Kg)*100=0.0427%`

I hope it helps!