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Chapter 4: Problem 67

(a) How many grams of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) , should you dissolve in water to make 1.00 \(\mathrm{L}\) of vodka (which is an aqueous solution that is 6.86 \(\mathrm{M}\) ethanol)? (b) Using the density of ethanol \((0.789 \mathrm{g} / \mathrm{mL}),\) calculate the volume of ethanol you need to make 1.00 \(\mathrm{L}\) of vodka.

Short Answer

Expert verified
(a) To make 1.00 L of vodka with a 6.86 M concentration, you need to dissolve 315.88 grams of ethanol in water. (b) To make 1.00 L of vodka, you need to add 400.36 mL of ethanol.

Step by step solution

01

1. Determine the moles of ethanol needed in the solution

First, we need to find the moles of ethanol we require to make 1.00 L of vodka with a 6.86 M concentration. The formula for molarity (M) is: M = moles of solute / volume of solution (in liters) We are given the molarity and volume, so we can solve for moles: moles = M × volume moles = 6.86 M × 1.00 L moles = 6.86 mol
02

2. Calculate the grams of ethanol needed

Now we need to convert the moles of ethanol into grams. The molar mass of ethanol can be found by adding the individual molar masses of its elements: Ethanol: \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\) Carbon: 12.01 g/mol (2 atoms) Hydrogen: 1.01 g/mol (6 atoms) Oxygen: 16.00 g/mol (1 atom) Molar mass of ethanol = (2 × 12.01) + (6 × 1.01) + (1 × 16.00) = 46.07 g/mol To find the grams of ethanol we need, we'll multiply the moles of ethanol by its molar mass: grams of ethanol = moles × molar mass grams of ethanol = 6.86 mol × 46.07 g/mol
03

3. Calculate the volume of ethanol required

We are given the density of ethanol (0.789 g/mL) and are asked to calculate the volume in milliliters. To find the volume, we'll use the formula: volume = mass / density volume = grams of ethanol / (0.789 g/mL) The final output will be our answers for parts (a) and (b). Solution:
04

1. Determine the moles of ethanol needed in the solution

moles = 6.86 M × 1.00 L moles = 6.86 mol
05

2. Calculate the grams of ethanol needed

Molar mass of ethanol = 46.07 g/mol grams of ethanol = 6.86 mol × 46.07 g/mol grams of ethanol = 315.88 g So, to make 1.00 L of vodka with a 6.86 M concentration, you need to dissolve 315.88 grams of ethanol in water.
06

3. Calculate the volume of ethanol required

volume = 315.88 g / (0.789 g/mL) volume = 400.36 mL So, to make 1.00 L of vodka, you need to add 400.36 mL of ethanol.

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Most popular questions from this chapter

The arsenic in a \(1.22-\) g sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of As in AsO \(_{4}^{3-2}(\mathbf{b})\) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took 25.0 \(\mathrm{mL}\) of 0.102 \(\mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

An aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of NaCl of the same concentration. Which of the following substances could the unknown be: KOH, \(\mathrm{NH}_{3}, \mathrm{HNO}_{3}, \mathrm{KClO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{3}, \mathrm{CH}_{3} \mathrm{COCH}_{3}(\) acetone)?

(a) Is the concentration of a solution an intensive or an extensive property? (b) What is the difference between 0.50 mol HCl and 0.50\(M \mathrm{HCl}\) ?

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) \(\mathrm{Cl}^{-},(\mathbf{b}) \mathrm{NO}_{3}^{-},(\mathbf{c}) \mathrm{NH}_{4}^{+},(\mathbf{d}) \mathrm{S}^{2}\) , (e) \(\mathrm{SO}_{4}^{2-} .\) [Section 4.2\(]\)

(a) Which will have the highest concentration of potassium ion: \(0.20 M \mathrm{KCl}, 0.15 M \mathrm{K}_{2} \mathrm{CrO}_{4},\) or 0.080\(M \mathrm{K}_{3} \mathrm{PO}_{4} ?\) (b) Which will contain the greater number of moles of potassium ion: 30.0 \(\mathrm{mL}\) of 0.15 \(\mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) or 25.0 \(\mathrm{mL}\) of 0.080 \(\mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4} ?\)
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