Answer:
The approximate molar mass of lauryl alcohol is 174.08 g/m
Explanation:
An excersise to apply the colligative property of Freezing-point depression.
This is the formula: ΔT = Kf . m
First of all, think the T° of fusion of benzene → 5.5°C
ΔT = T° pure solvent - T° fusion solution
Kf for benzene: 5.12 °C/m
5.5°C - 4.5°C = 5.12 °C /m . m
1°C / 5.12 m /°C = m
0.195 m = molality
This moles of lauryl alcohol, solute, are in 1 kg of benzene, solvent.
I have to find out in 0.2 kg.
1 kg sv ____ 0.195 moles solute
0.2 kg sv ____ (0.195 . 0.2)/1 = 0.039 moles solute
The mass for these moles is 6.80 g, so if I want to know the molar mass, I have to divide mass / moles
6.80 g/ 0.039 moles = 174.08 g/m
=
. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.