Question

Benzaldehyde ( = 106.1 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? Kf = 1.99°C/m, freezing point of pure ethanol = –117.3°C.

Answer 1

Answer:

The question to your answer is: Tc = -118.97°C

Explanation:

data

MW Benzaldehyde = 106.1 g/mol

Tc = ?

Mass Benzaldehyde = 75 g

Mass ethanol = 850 g

Kf = 1.99

Freezing point of ethanol = -117.3°C

Formula

                -Tc + Tc (solvent) = Kcm

molality

            106.1 g ------------------- 1 mol

              75 g  -------------------  x

    x = 75 / 106.1 = 0.71 mol

molality = #moles / kg solvent = 0.71 / 0.85 = 0.84

Substitution

                     -Tc + Tc (solvent) = Kcm

                     -Tc -117.3 = (1.99)(0.84)

                     -Tc = 117.3 + 1.67

                    - Tc = 118.97°C

                       Tc = -118.97°C

Answer 2

The freezing point of the solution is -118.97°C.

What are colligative properties?

The term colligative properties refers to those properties that depend on the number of molecules present.

Number of moles of solute = 75.00 g/ 106.1 g/mol = 0.71 moles

Molality of the solution = 0.71 moles/850.0 * 10^-3 = 0.84 m

Given that;

ΔT = K m i

ΔT = 1.99°C/m * 0.84 m * 1

ΔT = 1.67°C

Freezing point of solution = Freezing point of pure solution - ΔT

–117.3°C  - 1.67°C = -118.97°C

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