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What is the change in the freezing point of water when 35.5 g of sucrose is dissolved in 55.0g of water?

Kt of water = -1.86°C/mol
Molar mass sucrose = 342.30 g/mol
j value of sugar = 1

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Answer:

To calculate the change in the freezing point of water when sucrose is dissolved in it, you can use the formula for freezing point depression, which is given by:

ΔTf = Kf * m

Where:

ΔTf is the change in freezing point.

Kf is the cryoscopic constant for the solvent (water in this case).

m is the molality of the solution.

First, calculate the molality (m) of the solution, which is the number of moles of solute (sucrose) divided by the mass of the solvent (water) in kilograms:

Calculate the number of moles of sucrose:

Moles of sucrose = Mass of sucrose / Molar mass of sucrose

Moles of sucrose = 35.5 g / 342.30 g/mol ≈ 0.1037 mol

Convert the mass of water to kilograms:

Mass of water = 55.0 g = 0.0550 kg

Calculate the molality (m):

Molality (m) = Moles of sucrose / Mass of water (in kg)

Molality (m) = 0.1037 mol / 0.0550 kg ≈ 1.8855 mol/kg

Now that you have the molality (m), you can calculate the change in freezing point (ΔTf) using the given cryoscopic constant (Kf) for water:

ΔTf = Kf * m

ΔTf = (-1.86°C/mol) * (1.8855 mol/kg)

ΔTf ≈ -3.51°C

So, the change in the freezing point of water when 35.5 g of sucrose is dissolved in 55.0 g of water is approximately -3.51°C.

Step-by-step explanation:

Have great day!

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