Answer:
The molar mass of lauryl alcohol is approximately 180 g/mol
Explanation:
Step 1: Data given
Mass of lauryl alcohol = 8.80 grams
Mass of benzene = 0.100 kg = 100 grams
Freezing point of Benzene is 5.5 °C
Kf value for benzene = 5.12 °C/molal
Step 2: Calculate the freezing point depression
ΔTf = 5.5 - 3.0 °C = 2.5 °C
Step 3: Calculate molality
⇒ with ΔTf = the freezing point depression = 2.5 °C
⇒ with i = the van't Hoff factor = 1
⇒ with kf = the free point depression constat pf benzene = 5.12°C/m
⇒ with m =the molality = TO BE DETERMINED
molality = ΔTf / kf
molality = 2.5 °C / 5.12 °C /m
molality = 0.488 molal
Step 4: Calculate moles lauryl alcohol
Molality = moles lauryl alcohol / mass benzene
Moles lauryl alcohol = molality * mass benzene
Moles lauryl alcohol = 0.488 m * 0.100 kg
Moles lauryl alcohol = 0.0488 moles
Step 5: Calculate molar mass of lauryl alcohol
Molar mass lauryl alcohol = mass lauryl alcohol/ moles lauryl alcohol
Molar mass lauryl alcohol = 8.80 grams / 0.0488 moles
Molar mass lauryl alcohol = 180 g/mol
The molar mass of lauryl alcohol is approximately 180 g/mol
