An alcohol, Ethanol (C2H5OH), is present in 1L of wine that is 12% alcohol by volume. The density of the solution is 0.984 g/mL, and the density of pure alcohol is 0.789 g/mL. (C2H5OH = 46.07 g/mol)
Answer these items 1-2 for part 2:
1. What is the weight of the alcohol present in wine?
2. What is the molarity of the solution?
weight of the solution is 0.984 g/mL x 1,000 mL = 984 grams
That 1,000 mL contains 12 mL EtOH/100 mL or 120 mL EtOH in 1,000 mL.
mass = volume x density = 120 mL x 0.789 g/mL = 94.7 g EtOH in 1 L of the wine solution.
M = mols/L. mols EtOH = g/molar mass = 94.7 g/46,1 = 2.05 mols.
That's 2.05 mols/L of solution.
To calculate the weight of the alcohol present in wine, we need to use the formula:
Weight of alcohol = Volume of solution x Density of solution x % Alcohol by volume
Given:
Volume of solution = 1 L
Density of solution = 0.984 g/mL
% Alcohol by volume = 12%
1. Weight of alcohol present in wine:
Weight of alcohol = 1 L x 0.984 g/mL x 12% = 0.118 g
Therefore, the weight of the alcohol present in wine is 0.118 g.
Now, to calculate the molarity of the solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in L)
Given:
Volume of solution = 1 L
Weight of alcohol = 0.118 g
Molar mass of ethanol (C2H5OH) = 46.07 g/mol
2. Molarity of the solution:
First, we need to convert the weight of alcohol to moles using the molar mass:
moles = Weight / Molar mass = 0.118 g / 46.07 g/mol = 0.002559 mol
Now, we can calculate the molarity:
Molarity (M) = 0.002559 mol / 1 L = 0.002559 M
Therefore, the molarity of the solution is 0.002559 M.
To find the weight of the alcohol present in wine, you need to know the volume of the wine and the alcohol content by volume. Here's how to calculate it:
Step 1: Calculate the volume of alcohol present in the wine.
To do this, multiply the volume of the wine by the alcohol content by volume (ABV).
Volume of alcohol = Volume of wine * ABV
In this case, the volume of the wine is given as 1 liter and the ABV is 12% (or 0.12 in decimal form).
Volume of alcohol = 1 L * 0.12 = 0.12 L
Step 2: Convert the volume of alcohol to grams using the density of the solution.
Density = Mass / Volume
Rearrange the formula to solve for mass:
Mass = Density * Volume
The density of the solution is given as 0.984 g/mL, so we need to convert the volume of alcohol to milliliters.
0.12 L * 1000 mL/L = 120 mL
Mass of alcohol = 0.984 g/mL * 120 mL = 118.08 g
Therefore, the weight of the alcohol present in wine is 118.08 grams.
To find the molarity of the solution, you need to know the number of moles of the solute (alcohol) and the volume of the solution in liters. Here's how to calculate it:
Step 1: Calculate the number of moles of alcohol.
To do this, divide the weight of the alcohol by its molar mass.
Molar mass of ethanol (C2H5OH) = 2 * atomic mass of C + 6 * atomic mass of H + atomic mass of O = 2 * 12.01 g/mol + 6 * 1.008 g/mol + 16.00 g/mol = 46.07 g/mol
Number of moles of alcohol = Mass of alcohol / Molar mass of alcohol = 118.08 g / 46.07 g/mol ≈ 2.561 mol
Step 2: Calculate the molarity of the solution.
To do this, divide the number of moles of alcohol by the volume of the solution in liters.
Volume of the solution is given as 1 L.
Molarity = Number of moles of alcohol / Volume of solution = 2.561 mol / 1 L = 2.561 M
Therefore, the molarity of the solution is 2.561 M.