105. If a pharmacist incorporates 1 pint of Alcohol USP into 5 L of a mouthwash formula that was initially labeled as 10% V/V, what would be the percentage of alcohol present in the resulting mixture?
a- 14.34 %
b- 17.34 %
c- 19.68 %
d- 22.43 %
e- 24.40 %
Solution:
First of all, let's know something about Alcohol USP;
Alcohol USP (United States Pharmacopeia) refers to a specific grade of alcohol that is used for medicinal purposes, such as for use as an antiseptic or as a solvent in the manufacturing of pharmaceuticals. The concentration of alcohol in Alcohol USP is typically 95% ethanol by volume, which means that it contains 95% ethanol and 5% water.
So, Alcohol USP is 95% Alcohol.
In the question it is 1 pint of Alcohol USP, so we need to find out the amount of alcohol in 1 pint of Alcohol USP;
1 pint = 473.18 ml
So, (95ml / 100ml) x 473.18 ml = 449.521 ml
Thus, 1 pint of Alcohol USP contains 449.521 ml of alcohol.
Now we need to find out the amount of alcohol in 5 L of 10% V/V;
10% V/V means 10 ml alcohol is there in 100 ml of solution;
so, in 5 L (5000ml) of solution, how much alcohol?
(10 ml / 100 ml) x 5000 ml = 500 ml
so 500 ml of alcohol is there in 5 L of 10% V/V solution.
After adding 1 pint of Alcohol USP into 5 L of 10% V/V formula;
The total volume of alcohol becomes;
449.521 ml + 500 ml = 949.521 ml
The total volume of the mixture becomes;
473.18 ml + 5000 ml = 5473.18 ml
Now lets calculate the percentage of alcohol present in the resulting mixture;
( 949.521 ml / 5473.18 ml ) x 100 = 17.34%
Therefore, the correct answer is b.
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