Question

5. The alcohol content of hard liquor is normally given in terms of the "proof," which is defined as twice the percentage by volume of ethanol (CH3OH) present. Calculate the number of grams of alcohol present in 1.00 L of 75-proof gin. The density of ethanol is 0.798 g/mL. (4 points)

Answer 1

Answer:

299 g of ethanol presents in 1,00 L of 75-proof gin

Explanation:

75-proof gin in 1,00L have a concentration in volume of ethanol of 75/2 = 37,5 %. This means:

$\frac{0,375 L ethanol}{1,00 L}$

0,375 L ≡ 375 mL

375 mL ethanol ×$\frac{0,789 g}{1 mL}$ = 299 g of ethanol presents in 1,00 L of 75-proof gin

I hope it helps!

Answer 2

Alcohol is an organic compound whose hydroxyl group is attached to the carbon atom. The number of grams of alcohol present in 1.00 L of 75-proof gin is 299 g.

What is alcohol?

Alcohol is an organic compound referred to as primary alcohol, ethanol.

It is used in drugs and mainly for drinking purposes.

Given,

The density of ethanol is 0.798 g/mL.

The volume of ethanol

75-proof gin in 1.00 L has a concentration in the volume of ethanol that is

$\dfrac{75}{2} =37.5 %$

0.375 l = 375ml

$375 \times \dfrac{0.789}{1\;ml} = 299\;g$

Thus, the number of grams of alcohol present is 299 g.

Learn more about alcohol, here:

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0,375 L ≡ 375 mL

375 mL ethanol × = 299 g of ethanol presents in 1,00 L of 75-proof gin