Acetic acid is a polar molecule and can form hydrogen bonds with molecules. Therefore, it has a high soulubility in water. Yet acetic acid is also soluble in benzene (C6H6), a nonpolar solvent that lacks the ability to form hydrogen bonds. A solution of 3.8g of CH2COOH in 80 g C6H6 has a freezing point of 3.5 degrees C. Calculate the molality of the solution
delta T = Kf*m.
You will need the freezing point of benzene IF that is 3.5 degrees and not 3.5 depression.
To calculate the molality of a solution, we need to first convert the given masses of solute and solvent into moles.
Step 1: Calculate the moles of acetic acid (CH2COOH):
The molar mass of acetic acid (CH2COOH) is:
12.01 g/mol (C) + 2.01 g/mol (H) + 32.04 g/mol (O) + 16.00 g/mol (O) = 60.05 g/mol
Given mass of CH2COOH = 3.8 g
Moles of CH2COOH = 3.8 g / 60.05 g/mol = 0.0633 mol
Step 2: Calculate the moles of benzene (C6H6):
The molar mass of benzene (C6H6) is:
12.01 g/mol (C) + 1.01 g/mol (H) = 78.11 g/mol
Given mass of C6H6 = 80 g
Moles of C6H6 = 80 g / 78.11 g/mol = 1.023 mol
Step 3: Calculate the molality:
Molality (m) is defined as moles of solute per kilogram of solvent.
Given mass of C6H6 = 80 g = 0.080 kg
Molality (m) = moles of CH2COOH / mass of C6H6 (kg)
m = 0.0633 mol / 0.080 kg = 0.791 mol/kg
Therefore, the molality of the solution is 0.791 mol/kg.
To calculate the molality of the solution, we need to first find the moles of acetic acid (CH2COOH) and the mass of benzene (C6H6).
Step 1: Find the moles of acetic acid (CH2COOH)
The molar mass of acetic acid (CH2COOH) can be calculated by adding up the atomic masses of each element in the molecule.
CH2COOH = (1 x 2) + 12 + (16 x 2) + 1 + 16 = 60 g/mol
To find the moles of acetic acid, we divide the mass of acetic acid (3.8g) by its molar mass (60 g/mol):
moles of acetic acid = 3.8g / 60 g/mol
Step 2: Find the mass of benzene (C6H6)
The mass of benzene is given as 80g.
Step 3: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we are given in grams, we need to convert the mass of benzene to kilograms.
mass of benzene (kg) = 80g / 1000 = 0.08 kg
Now, we can calculate the molality (m) using the formula:
m = moles of solute / mass of solvent (in kg)
m = (moles of acetic acid) / (mass of benzene in kg)
Plug in the values we calculated in Step 1 and Step 3:
m = (3.8g / 60 g/mol) / 0.08 kg
Simplify and calculate:
m = 0.0633 mol / 0.08 kg
m ≈ 0.7913 mol/kg
Therefore, the molality of the solution is approximately 0.7913 mol/kg.