Pure glacial acetic acid HC2H3O2 has a freezing point of 16.62 degrees C. Its freezing point depression constant is Kf=3.57 Cm^-1. A solution was made by taking 9.755 g of an unknown non-electrolyte and dissolving it in 90.50 g of glacial acetic acid. The measured freezing point of the solution was 8.64 degrees C. Find the molecular weight of the unknown substance.

Question

Pure glacial acetic acid HC2H3O2 has a freezing point of 16.62 degrees C. Its freezing point depression constant is Kf=3.57 Cm^-1. A solution was made by taking 9.755 g of an unknown non-electrolyte and dissolving it in 90.50 g of glacial acetic acid. The measured freezing point of the solution was 8.64 degrees C. Find the molecular weight of the unknown substance.

Answer 1

To find the molecular weight of the unknown substance, we can use the formula for freezing point depression:

ΔT = Kf * m

Where:
ΔT is the change in freezing point (in this case, the freezing point of the pure solvent minus the freezing point of the solution)
Kf is the freezing point depression constant
m is the molality of the solution

First, let's calculate the molality of the solution:

Molality (m) = moles of solute / mass of solvent (in kg)

Given:
Mass of glacial acetic acid (solvent) = 90.50 g
Mass of unknown non-electrolyte (solute) = 9.755 g

First, convert the masses to kilograms:

Mass of glacial acetic acid = 90.50 g ÷ 1000 = 0.09050 kg
Mass of unknown non-electrolyte = 9.755 g ÷ 1000 = 0.009755 kg

Now we can calculate the molality:

Molality (m) = 0.009755 kg / 0.09050 kg = 0.1077 mol/kg

Next, we can calculate the change in freezing point (ΔT):

ΔT = Freezing point of pure solvent - Freezing point of the solution

Given:
Freezing point of pure glacial acetic acid = 16.62 °C
Freezing point of the solution = 8.64 °C

ΔT = 16.62 °C - 8.64 °C = 7.98 °C

Finally, we can solve for the molecular weight (M) of the unknown non-electrolyte using the formula:

ΔT = Kf * m
7.98 °C = 3.57 °C·kg/mol * 0.1077 mol/kg

Rearranging the equation, we get:

M = ΔT / (Kf * m)
M = 7.98 °C / (3.57 °C·kg/mol * 0.1077 mol/kg)

Calculating M:

M = 7.98 °C / (3.57 °C·kg/mol * 0.1077 mol/kg)
M = 7.98 °C / 0.385 kg/mol
M = 20.7 g/mol

Therefore, the molecular weight of the unknown substance is approximately 20.7 g/mol.

Answer 2

To find the molecular weight of the unknown substance, we can use the formula for freezing point depression:

ΔT = Kf * m

where:
ΔT = freezing point depression
Kf = freezing point depression constant
m = molality of the solution

First, let's calculate the molality of the solution:

mass of solute = 9.755 g
mass of solvent = 90.50 g (glacial acetic acid)

First, convert the mass of the solvent to moles:
moles of glacial acetic acid = mass / molar mass
molar mass of glacial acetic acid (CH3COOH) = 60.052 g/mol
moles of glacial acetic acid = 90.50 g / 60.052 g/mol

Next, calculate the molality of the solution:
molality (m) = moles of solute / mass of solvent (in kg)
mass of solvent = 90.50 g / 1000 g/kg
molality = moles of solute / 0.09050 kg

Now, let's calculate the freezing point depression:

ΔT = Kf * m
ΔT = 8.64°C - 16.62°C
ΔT = -7.98°C

Kf is given as 3.57°C/m

-7.98°C = 3.57°C/m * (moles of solute / 0.09050 kg)

Rearranging the equation to solve for moles of solute:

moles of solute = (-7.98°C * 0.09050 kg) / 3.57°C/m

Finally, let's calculate the molecular weight of the unknown substance:

molecular weight = mass of solute / moles of solute

Remember, the mass of the solute provided was 9.755 g.

molecular weight = 9.755 g / moles of solute

Plug in the value calculated for moles of solute to get the molecular weight of the unknown substance.

Answer 3

depression=molality*Kf

= gramsunknown*Kf/(molmassunk*.0905)

solve for molmassunk.