**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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