Question

Benzaldehyde (106.12 g/mol), also known as oil of almonds, is used in the manufacture of dyes and perfumes and in flavorings. What would be the freezing point of a solution prepared by dissolving 75.00 g of benzaldehyde in 850.0 g of ethanol? (For ethanol, Kf = 1.99°C/m; freezing point of pure ethanol = –117.3°C)

Answer 1

T° freezing solution = -118.9 °C

Step-by-step explanation:

This is about colligative property of freezing point depression

ΔT = Kf . m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryoscopic constant

m = molality (mol of solute in 1kg of solvent)

Let's convert mass of solute to moles ( mass / molar mass)

75 g / 106.12 g/m = 0.707 moles

These are the moles of solute contained in 850 g of solvent. We must find out the moles that occupy 1 kg

0.707 moles / 0.850 kg = 0.83 molality

T° freezing pure solvent - T° freezing solution = 1.99°C/m . 0.83 m

-117.3°C - T° freezing solution = 1.99°C/m . 0.83 m

T° freezing solution = 1.99°C/m . 0.83 m + 117.3°C

T° freezing solution = -118.9 °C