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)

T° freezing solution =  -118.9 °C

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

h let me think

2b2t

explanation:

2b2t