Question

Camphor (C10H16O) melts at 179.8°C, and it has a particularly large freezing-point-depression constant, Kf= 40.0ºC/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7°C. What is the molar mass of the solute?

Answer 1

The molar mass of the organic compound = 109.04 g/mol

Step-by-step explanation:

Step 1: Data given

Camphor (C10H16O) melts at 179.8°C

freezing-point-depression constant, Kf= 40.0ºC/m

Mass of organic substance = 0.186 grams

Mass of camphor = 22.01 grams

The freezing point of the mixture = 176.7°C

Step 2: Calculate molality

ΔT = i*kf * m

⇒ ΔT = The freezing point depression = 179.8 - 176.7 = 3.1 °C

⇒ i = the van't Hoff factor = 1

⇒ kf = the freezing point depression constant = 40.0 °C/m

⇒ m = the molality = moles organic substance / mass camphor

3.1 °C = 1* 40.0 °C/m * m

m = 3.1 °C / 40.0 °C/m

m = 0.0775 molal

Step 3: Calculate moles

molality = moles organic substance / mass camphor

0.0775 molal = moles organic substance / 0.02201 kg

moles organic substance = 0.0775 * 0.02201

moles organic substance = 0.001705775 moles

Step 4: Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.186 grams / 0.001705775

Molar mass of the organic compound = 109.04 g/mol

The molar mass of the organic compound = 109.04 g/mol

Answer 2

Molar mass for the unknown solute is 109 g/mol

Step-by-step explanation:

Freezing point depression is the colligative property that must be applied to solve the question.

T°F pure solvent - T°F solution = Kf . m

Let's analyse the data given

Camphor → solvent

Unknown solute → The mass we used is 0.186g

T°F pure solvent = 179.8°C and T°F solution = 176.7°C.

These data help us to determine the ΔT → 179.8°C - 176.7°C = 3.1°C

So we can replace → 3.1°C = 40°C/m . m

m = 3.1°C / 40 m/°C → 0.0775 mol/kg

We have these moles of solute in 1kg of solvent, but our mass of camphor is 22.01 g (0.02201 kg).

We can determine the moles of solute → molality . kg

0.0775 mol/kg . 0.02201 kg = 1.70×10⁻³ moles

Molar mass → mass (g) / moles → 0.186 g / 1.70×10⁻³ mol = 109 g/mol