Answer to Question #313129 in Chemistry for KTV

Question

a solution of 3.80 g of CH3COOH in 80.0 g C6H6 has a freezing point of 3.50°C. calculate molar mass of the solute (CH3COOH). the normal freezing point of pure benzene is 5.50°C

Answer 1

Solution:

The lowering (depression) of the freezing point of the solvent can be represented using the following equation: 

Δt = i × K_f × m

where:

Δt = the change in freezing point

i = van't Hoff factor

K_f = the freezing point depression constant

m = the molality of the solute

 

The normal freezing point of pure benzene is 5.50°C

Therefore, Δt = 5.50°C − 3.50°C = 2.00°C

Typically, ions do not form in a nonpolar solvent.

Therefore, i = 1 (since acetic acid is dissolving in a nonpolar solvent - benzene)

K_f = 5.12 °C kg mol¯¹ (for benzene)

Thus,

m = Δt / (i × K_f)

Molality of CH₃COOH solution = (2.00°C) / (1 × 5.12 °C kg mol¯¹) = 0.39 mol kg¯¹

Molality = Moles of solute / Kilograms of solvent

Kilograms of solvent = Kilograms of benzene = (80.0 g) × (1 kg / 1000 g) = 0.08 kg

Therefore,

Moles of CH₃COOH = Molality of CH₃COOH solution × Kilograms of benzene

Moles of CH₃COOH = (0.39 mol kg¯¹) × (0.08 kg) = 0.0312 mol

Moles = Mass / Molar mass

Therefore,

Molar mass of CH₃COOH = Mass of CH₃COOH / Moles of CH₃COOH

Molar mass of CH₃COOH = (3.80 g) / (0.0312 mol) = 121.8 g/mol

The molar mass of the solute (CH₃COOH) is 121.8 g/mol

The calculated answer is approximately double the known molar mass (60.0 g/mol) of acetic acid. What acetic acid does in benzene is to form dimers, composed of two molecules of acetic acid chemically joined together.

Acetic acid dimer formula = (CH₃COOH)₂