Question

1. Calculate the molarity of a solution containing 49.3g of dissolved naphthalene (C10H8) in 217g benzene (C6H6).

2. A solution is prepared by mixing 94.2g of H2O, and 58.7g of ethanol, C2H5OH. Determine the mole fractions of each substance.

3. If you use 2.17mol of sucrose (C12H22O12) and dissolve this into 0.500kg of H2O, what will be the change in the freezing point of your solution? The freezing point depression of water (Kf) is 1.86 degrees Celsius/m.

Answer 1

1. 1.56 M

2. Water → 0.803

Ethanol → 0.197

3. - 8.07°C

Step-by-step explanation:

1. To solve this problem, we need benzene's density in order to determine the volume.

We assume the volume of solvent, as the volume of solution

Density for benzene is 0.8786 g/mL.

Density = m/V → V = m/density

Volume = 217 g /0.8786 g/mL = 247 mL

We convert the mass to moles and then, to mmoles

49.3 g . 1mol / 128g . 1000mmol / mol = 385.1 mmol

M = 385.1 mmol / 247mL = 1.56 M

2. We determine moles of each, solute and solvent

94.2 g. 1mol /18g = 5.23 moles of water

58.7 g . 1mol /46g = 1.28 moles of ethanol.

Total moles = 5.23 + 1.28 = 6.51

Mole fraction of water = Moles of water / Total moles

Mole fraction of ethanol = Moles of ethanol / Total moles

Water → 5.23 / 6.51 = 0.803

Ethanol → 1.28 / 6.51 = 0.197

3. Freezing point depression → ΔT = Kf . m . i

i = Van't Hoff factor, ions that are present in the solution.

As the solute is organic, i = 1.

Kf = 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

2.17 mol / 0.5kg = 4.34 m

ΔT = Freezing T° of pure solvent - Freezing T° of solution.

We replace data → 0° - Freezing T° of solution = 4.34m . 1.86°c/m . 1

Freezing T° of solution = - 8.07°C