The mole fraction of acetic acid (CH3COOH) in an aqueous solution is 0.675. Calculate the molarity of the acetic acid, if the density of the solution is 1.0266 g mL-1.
To calculate the molarity of acetic acid, we first need to convert the given mole fraction to the concentration in terms of moles per liter.
The mole fraction (X) is defined as the ratio of moles of the solute to the total moles of the solution. In this case, we have the mole fraction of acetic acid (CH3COOH) in the aqueous solution as 0.675.
To calculate the molarity, we need to know the total volume of the solution. However, the density of the solution is given instead. We can use the density to convert the mass of the solution to its volume.
First, we need to determine the total mass of the solution. The density is given as 1.0266 g mL-1. Let's assume we have 1 liter of solution (since molarity is defined as moles per liter):
Total mass of the solution = density × volume
= 1.0266 g mL-1 × 1000 mL
= 1026.6 g
Now, we can calculate the mass of acetic acid in the solution using the mole fraction:
Mass of acetic acid = mole fraction × total mass of the solution
= 0.675 × 1026.6 g
= 692.535 g
Next, we need to calculate the moles of acetic acid present in the solution using its molar mass. The molar mass of acetic acid (CH3COOH) is:
Molar mass of acetic acid (CH3COOH) = (12.01 g/mol × 2) + (1.01 g/mol × 4) + (16.00 g/mol × 2)
= 60.05 g/mol
Moles of acetic acid = mass of acetic acid / molar mass of acetic acid
= 692.535 g / 60.05 g/mol
≈ 11.53 mol
Finally, we can calculate the molarity of acetic acid:
Molarity (M) = moles of acetic acid / volume of solution in liters
Since we assumed 1 liter of solution, the molarity is equal to the number of moles:
Molarity (M) = 11.53 mol/L
Therefore, the molarity of acetic acid in the aqueous solution is approximately 11.53 M.
To calculate the molarity (M) of acetic acid in the solution, we need to know the molecular weight (MW) of acetic acid.
The molecular weight of acetic acid is:
C: 12.01 g/mol
H: 1.01 g/mol (3 atoms)
O: 16.00 g/mol
MW (acetic acid) = (12.01 * 2) + (1.01 * 4) + 16.00 = 60.05 g/mol
Next, we need to convert the density of the solution into grams per milliliter (g/mL) to calculate the number of moles of acetic acid.
Given:
Density (solution) = 1.0266 g/mL
To find the mass of the solution, we multiply the density by the volume. Since the volume is not given, we assume it to be 1 L (1000 mL):
Mass (solution) = 1.0266 g/mL * 1000 mL = 1026.6 g
Now, we can calculate the number of moles of acetic acid present in the solution:
Moles (acetic acid) = mole fraction * moles (solution)
The mole fraction of acetic acid (CH3COOH) is given as 0.675.
Moles (acetic acid) = 0.675 * Moles (solution)
Moles (acetic acid) = 0.675 * (Mass (solution) / MW (acetic acid))
Moles (acetic acid) = 0.675 * (1026.6 g / 60.05 g/mol)
Moles (acetic acid) = 11.528 mol
Finally, we can calculate the molarity of acetic acid by dividing the moles of acetic acid by the volume of the solution in liters:
Molarity (acetic acid) = Moles (acetic acid) / Volume (solution)
Since the volume of the solution is assumed to be 1 L:
Molarity (acetic acid) = 11.528 mol / 1 L
Molarity (acetic acid) = 11.528 M
Therefore, the molarity of the acetic acid in the solution is 11.528 M.