Respuesta :
Answer:
Molar mass for the unknown solute is 109 g/mol
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
Answer:
The molar mass of the organic compound = 109.04 g/mol
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
