Question

(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

(a) To make 1.00 L of vodka, you need to dissolve 315.92 grams of ethanol, CH3CH2OH, in water. (b) Using the density of ethanol (0.789 g/mL), you need 400.51 mL of ethanol to make 1.00 L of vodka.

Step-by-step solution

Identify the given information and desired outcomes

We are given: (a) Molarity (M) of ethanol in vodka: 6.86 M (b) Density of ethanol: 0.789 g/mL We need to find: (a) Grams of ethanol required to make 1 L of vodka (b) Volume of ethanol required to make 1 L of vodka

Calculation of moles of ethanol needed

Since we know the molarity and volume of the solution, we can calculate the moles of ethanol required. Molarity is defined as moles per liter, so we can multiply the volume with the molarity to calculate the total moles of ethanol needed: Moles of ethanol = Molarity × Volume Moles of ethanol = 6.86 mol/L × 1 L Moles of ethanol = 6.86 mol

Calculation of grams of ethanol needed

Now that we have calculated the moles of ethanol needed, we can convert this to grams of ethanol using the molecular weight of ethanol, which is: Molecular weight of ethanol, CH3CH2OH = 12.01 g/mol (C) + 1.01 g/mol (6H) + 16.00 g/mol (O) Molecular weight of ethanol = 46.07 g/mol Using the moles of ethanol and molecular weight, we can calculate the grams of ethanol: Grams of ethanol = Moles × Molecular weight Grams of ethanol = 6.86 mol × 46.07 g/mol Grams of ethanol = 315.92 g Therefore, 315.92 grams of ethanol is required to make 1 L of vodka.

Calculation of the volume of ethanol needed

Now that we know the grams of ethanol required, we can calculate the volume using the given density: Volume of ethanol = Mass / Density Volume of ethanol = 315.92 g / (0.789 g/mL) Volume of ethanol = 400.51 mL Therefore, 400.51 mL of ethanol is required to make 1 L of vodka.