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Chapter 4: Problem 65

The concentration of alcohol $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)$ in blood, called the "blood alcohol concentration" or BAC, is given in units of grams of alcohol per \(100 \mathrm{~mL}\) of blood. The legal definition of intoxication, in many states of the United States, is that the BAC is 0.08 or higher. What is the concentration of alcohol, in terms of molarity, in blood if the BAC is \(0.08 ?\)

Short Answer

Expert verified
The concentration of alcohol in the blood when the BAC is 0.08 is approximately \(1.74 \times 10^{-5} \text{ M}\).

Step by step solution

01

Calculate the Molecular Weight of Alcohol

The molecular weight of alcohol can be calculated by adding up the weights of all the atoms in the molecule, which is: Molecular weight of alcohol = Molecular weight of C + Molecular weight of C + Molecular weight of H × 3 + Molecular weight of H × 2 + Molecular weight of O + Molecular weight of H - Molecular weight of C = 12.01 g/mol - Molecular weight of H = 1.01 g/mol - Molecular weight of O = 16.00 g/mol So, the molecular weight of alcohol is: \(12.01 + 12.01 + 3×1.01 + 2×1.01 + 16.00 + 1.01 = 46.07 \text{ g/mol}\)
02

Calculate the Mass of Alcohol in Blood

The given BAC value is 0.08, which means there are 0.08 grams of alcohol per 100 mL of blood. We need to convert the volume of blood to liters: \(0.08 \frac{\text{g}}{100 \text{ mL}} × \frac{1 \text{ L}}{1000 \text{ mL}} = 0.0008 \frac{\text{g}}{\text{L}}\)
03

Convert the Mass to Moles

Now, we can convert the mass of alcohol to moles using the molecular weight calculated in Step 1: \(\text{Moles of alcohol} = \frac{\text{Mass of alcohol}}{\text{Molecular weight of alcohol}}\) \(\text{Moles of alcohol} = \frac{0.0008 \text{ g}}{46.07 \text{ g/mol}} = 1.736×10^{-5} \text{ mol}\)
04

Calculate the Molarity of Alcohol in Blood

Finally, the molarity is the number of moles of solute (alcohol) per liter of solvent (blood). Since we already have calculated the moles of alcohol (1.736 × 10^{-5} mol) and the blood volume in liters (1 L), we can now calculate the molarity: Molarity = \(\frac{\text{Moles of alcohol}}{\text{Volume of blood in liters}}\) Molarity = \(\frac{1.736 \times 10^{-5} \text{ mol}}{1 \text{ L}} = 1.736 \times 10^{-5}\text{ M}\) Therefore, the concentration of alcohol in the blood when the BAC is 0.08 is approximately \(1.74 \times 10^{-5} \text{ M}\).

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Most popular questions from this chapter

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2},\) or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} . \mathrm{A}\) friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) Silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

The distinctive odor of vinegar is due to aceticacid, $\mathrm{CH}_{3} \mathrm{COOH}$, which reacts with sodium hydroxide according to: \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCH}_{3} \mathrm{COO}(a q) $$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of $0.115 \mathrm{M} \mathrm{NaOH}$ to reach the equivalence point in a titration, how many grams of acetic acid are in a \(1.00-\) qt sample of this vinegar?

State whether each of the following statements is true or false. Justify your answer in each case. (a) Sulfuric acid is a monoprotic acid. (b) \(\mathrm{HCl}\) is a weak acid. (c) Methanol is a base.

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing $1.00 \times 10^{2} \mathrm{~mL}\( of \)0.0105 \mathrm{MHCl} .\( The \)\mathrm{NH}_{3}$ reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

A \(4.36-g\) sample of an unknown alkali metal hydroxide is dissolved in $100.0 \mathrm{~mL}$ of water. An acid-base indicator is added, and the resulting solution is titrated with \(2.50 \mathrm{MHCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(17.0 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the alkali metal cation: $\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+},\( or \)\mathrm{Cs}^{+} ?$
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