Determine the freezing point of a solution which contains 0.31 mol of Sucrose in

Question

Determine the freezing point of a solution which contains 0.31 mol of Sucrose in 175 g of water. [For water, K_f = 1.86 degree C/m] Select one: 3.3 C 1.1 degree C 0.0 degree C -1.1 degree C -3.3 degree C 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, K_f = 1.99 degree C/m, freezing point of pure ethanol = -117.3 degree C] Select one: -117.5 degree C -118.7 degree C -119.0 degree C -120.6 degree C -121.1 degree C

Explanation / Answer

1)

Since freezing point depression = change in temperature

So, Change in T = -Kf x molality of solution

In this solution, molality = 0.31 mol sucrose / .175 kg water =1.8 moles sucrose / kg water.

Therefore,

Change in T = - 1.86 C/m (1.8 molal sucrose) = -3.3°C

water freezes at 0 C. But when sucrose added to the solution, the freezing point drops to about -3.3° C.

So, option (e) is the correct choice.

2)

moles benzaldehyde = 75g / 106.1g/mole = 0.7068 moles

Solvent ethanol = 0.85 Kg
So molarity = 0.7068 moles / 0.85kg = 0.831 m

T = Kf x m
T = 1.99ºC/m x 0.831 m = 1.653ºC
new freezing point = -117.3 - 1.651 = -118.953ºC
Thats nearly -119°C i.e choice (c) .

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