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

Respuesta :

Answer:

a. 1250 mL

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.

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