Question

1. Consider a solution of 1.058 g of eugenol (C10H12O2), an oil found in cloves, in 15.00 g of benzene (C6H6). Assume that eugenol is a nonvolatile non-electrolyte solute. The density of the mixture is 0.8871 g/mL. For benzene, Kb is 2.61ºC/m, Kf is 5.065 C/m, the vapor pressure at 38.4 ºC is 170.9 mm Hg, the melting (=freezing) point is 5.48 ºC and the boiling point is 80.10 ºC. (Refer to your text or lecture powerpoint for the necessary equations for vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure.) Calculate the following:

a. mass percent of solute in this solution .

b. molality of the solution.

c. molarity of the solution.

d. mole fraction of solute in the solution.

e. mole fraction of solvent in the solution

Answer 1

A. Mass percentage = mass of solute * 100 / mass of solution

= 1.058 *100 / ( 1.058 + 15.00 )

= 6.588%

B. Moles of eugenol = mass / molar mass

= 1.058 / 164.2 = 0.00644

Molality = moles / mass of solvent ( in kg )

= 0.00644 / ( 15.00*10^-3) = 0.43

C. Total mass of solution = 15.00 + 1.058 = 16.058 grams

Volume = mass / density

= 16.058 / 0.8871 = 18.1 ml

Molarity = moles / volume ( in L )

= 0.00644 / (18.1*10^-3)

= 0.356

D. Moles of benzene = 15 / 78 = 0.192

Mole fraction of solute = moles of solute / total moles

= 0.00644 / ( 0.192 + 0.00644)

= 0.03245

E. Mole fraction of solvent = 1 - mole fraction of solute

= 1 - 0.03245 = 0.96755

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