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How many milliliters of a 1:5000 w/v solution of the preservative benzalkonium chloride can be made from 125 mL of a 0.2% w/v solution of the preservative? a. 1250 mL b. 2500 mL C. 625 mL d. 1875 ml

User Carmenism
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2 Answers

3 votes

Final answer:

To determine the volume of a 1:5000 w/v solution that can be made from 125 mL of a 0.2% w/v solution of benzalkonium chloride, we calculate the mass of the solute in the initial solution and divide it by the concentration of the desired dilute solution, giving us 1250 mL.

Step-by-step explanation:

To calculate the target volume of a 1:5000 w/v solution from a 0.2% w/v solution of benzalkonium chloride, we first need to understand the concentration terms. A 0.2% w/v solution means 0.2 g of solute is present in every 100 mL of solution. For a 1:5000 w/v solution, there is 1 g of solute in 5000 mL of solution, or essentially y, 0.0002 g/mL (1 g / 5000 mL).

We then figure out how much benzalkonium chloride is in the initial 125 mL of the 0.2% solution. Since 0.2% w/v is equivalent to 0.2 g/100 mL, we have 0.2 g in 100 mL, hence in 125 mL, we'll have 0.25 g (since 125 mL is 1.25 times 100 mL).

To work out how much of the 1:5000 solution we can make with 0.25 g, we use the concentration of the 1:5000 solution, which is 0.0002 g/mL. Dividing the total available mass of solute (0.25 g) by the concentration (0.0002 g/mL), we get 1250 mL. Thus, the answer is a. 1250 mL.

User Edor Linus
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4 votes

Answer:

a. 1250 mL

Step-by-step explanation:

We use the C1V1=C2V2 equation to solve this problem, with:

C1=1/5000=0,0002 w/v, V1=?

C2=0,2%=0,2/100=0,002 w/v V2=125 mL

Then:

V1=C2V2/C1=0,002*125/0,0002=125O mL

Thus, 1250 mL of the preservative 1:5000 w/v can be made from 125 mL of a 0,2% w/v solution.

User Jonathan Sayce
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