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Paladinen [302]
3 years ago
9

How many grams of drug powder would you need to weigh to prepare 32ml of a suspension that should have a strength of 20mg/ml?

Chemistry
2 answers:
butalik [34]3 years ago
8 0
15 grams for the drug powder
Levart [38]3 years ago
3 0

Answer:

The answer is 0.64 g of drug powder.

Explanation:

Let "x" be the grams needed to prepare a 20 mg/mL mass concentration solution, let "V" be the volume of the solution, in this case 32 mL, therefore the mass concentration will be equal to:

mass concentration = x/V

Clearing x, we have:

x = mass concentration * V = (20 mg/mL) * (32 mL) = 640 mg

one gram is one thousand milligrams, therefore, we have:

640 mg * 1 g/1000 mg = 0.64 g

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2Al+6HBr\rightarrow 2AlBr_3+3H_2

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

8molAl(\frac{6molHBr}{2molAl})

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

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