Question

The molecular weight of ethanol (CHgCHzOH) is 46 and its density is 0.789 g/cm3. A. What is the molarity of ethanol in beer that is 5% ethanol by volume? [Alcohol content of beer varies from about 4Vo (lite beer) to B% (stout beer). B. The legal limit for a driver's blood alcohol content varies, but 80 mg of ethanol per 100 mL of blood (usually referred to as a blood alcohol level of 0.08) is typical. What is the molarity of ethanol in a person at this legal limit? C. How many l2-oz (355-mL) bottles of 5% beer could a 70-kg person drink and remain under the legal limit? A 70-kg person contains about 40 liters of water. Ignore the metabolism of ethanol, and assume that the water content of the person remains constant. D. Ethanol is metabolized at a constant rate of about 120 mg per hour per kg body weight, regardless of its concentration. If a 70-kg person were at twice the legal limit (160 mg/f 00 mL), how long would it take for their blood alcohol level to fall below the legal limit?

Answer 1

Solution.

A) Let's assume we have 1 L of beer. It contains 1*0.05 =0.05 L, or 50 cm³ of ethanol. It weights 0.789*50 = 39.45 grams. The amount of substance is n = m/M = 39.45/46 = 0.858 moles.

The concentration is c = n/V = 0.858/1 = 0.858 M.

B) 80 mg of ethanol per 100 mL of blood is the same as 800 mg per 1000 mL of blood, or 0.8 g/L. It equals to a molarity of c = n/V = 0.8/46 = 0.017 M.

C) The volume of solution is 40 L. The amount of ethanol in beer consumed and in human liquids is the same, so

The volume of beer should be consumed is

The number of bottles is 0.793/0.355 = 2.234, so one should drink less than 2 bottles to remain under the legal limit.

D) The rate of alcohol metabolism of this 70-kg person is 120*70 = 8400 mg/h. The initial amount of ethanol in a body is 1600*40 = 64000 mg. The legal limit is 800*40 = 32000 mg, so 64000-32000 = 32000 mg of ethanol should be removed. It reqires 32000/8400 = 3.81 hours.