What is the freezing point depression when 309 g of isoprene (C5H8) is dissolved in 747 g of ethyl ether? The freezing point constant for ethyl ether is 1.79 °C/m.3. What is the freezing point depression when 62.2 g of toluene (C7H8) is dissolved in 481 g of naphthalene? The freezing point constant for naphthalene is 7.00 °C/m.. Created by Eros01.

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Answer:

ΔTf = Kf x m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

For the first problem:

First, we need to calculate the molality of the solution:

Moles of isoprene = mass/molar mass = 309 g / 68.12 g/mol = 4.537 mol

Moles of ethyl ether = mass/molar mass = 747 g / 74.12 g/mol = 10.080 mol

Molality of the solution = moles of solute / mass of solvent (in kg)

= 4.537 mol / 0.747 kg = 6.07 mol/kg

Now, we can use the formula to calculate the freezing point depression:

ΔTf = Kf x m = 1.79 °C/m x 6.07 mol/kg = 10.87 °C

Therefore, the freezing point depression is 10.87 °C.

For the second problem:

First, we need to calculate the molality of the solution:

Moles of toluene = mass/molar mass = 62.2 g / 92.14 g/mol = 0.675 mol

Moles of naphthalene = mass/molar mass = 481 g / 128.17 g/mol = 3.753 mol

Molality of the solution = moles of solute / mass of solvent (in kg)

= 0.675 mol / 0.481 kg = 1.405 mol/kg

Now, we can use the formula to calculate the freezing point depression:

ΔTf = Kf x m = 7.00 °C/m x 1.405 mol/kg = 9.83 °C

Therefore, the freezing point depression is 9.83 °C.

Explanation:

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