Question

How many milliliters of a 2.5% w/v chlorpromazine hydrochloride injection and how many milliliters of 0.9% w/v sodium chloride injection should be used to prepare 500 mL of a 0.3% w/v chlorpromazine hydrochloride injection

Answer 1

Answer:

Therefore, 107 ml of 2.5% w/v chlorpromazine hydrochloride injection and 393 ml of 0.9% w/v sodium chloride injection are needed to prepare 500 mL of a 0.3% w/v chlorpromazine hydrochloride injection

Explanation:

The method to be used in determining the proportions of each component to be used to get the desired mixture is known as alligation alternate.

Alligation alternate is a method by which to calculate the number of parts of two  or more components of a given strength when they are to be mixed to a desired  strength. The difference between the strength of the stronger component and the  desired strength indicates the number of parts of the weaker component to be used and the difference between the desired strength and the strength of the weaker  component indicates the number of parts of the stronger to be used.

Desired strength = 0.3%

Number of parts of 2.5% w/v chlorpromazine hydrochloride injection to be used = 0.9 - 0.3 = 0.6 parts

Number of parts of 0.9% w/v sodium chloride injection to be used = 2.5 - 0.3 = 2.2 parts

Total parts = 2.8 parts

2.8 parts of 0.3% w/v chlorpromazine hydrochloride injection = 500 ml

0.6 parts of 2.5% w/v chlorpromazine hydrochloride injection = 0.6 * 500 ml/2.8 = 107 ml of 2.5% w/v chlorpromazine hydrochloride injection

2.2 parts of 0.9% w/v sodium chloride injection = 2.2 * 500 / 2.8 = 393 ml of 0.9% w/v sodium chloride injection

Therefore, 107 ml of 2.5% w/v chlorpromazine hydrochloride injection and 393 ml of 0.9% w/v sodium chloride injection are needed to prepare 500 mL of a 0.3% w/v chlorpromazine hydrochloride injection