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Chapter 13: Problem 42

A solution is made containing 20.8 \(\mathrm{g}\) of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) in 425 \(\mathrm{g}\) of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right) .\) Calculate (a) the mole fraction of phenol, ( b) the mass percent of phenol, (c) the molality of phenol.

Short Answer

Expert verified
The short answer is: a) Mole fraction of phenol = 0.0234 b) Mass percent of phenol = 4.67% c) Molality of phenol = 0.519 mol/kg

Step by step solution

01

Calculate the moles of phenol and ethanol

First, we need to find the moles of phenol and ethanol in the solution. We can do this using the given masses and the molar masses of the two compounds. Molar mass of phenol (C6H5OH) = 6(12.01) + 5(1.01) + 16.00 + 1.01 = 94.11 g/mol Moles of phenol = mass of phenol / molar mass of phenol Moles of phenol = 20.8 g / 94.11 g/mol = 0.221 mol Molar mass of ethanol (CH3CH2OH) = 12.01 + 3(1.01) + 12.01 + 2(1.01) + 16.00 + 1.01 = 46.07 g/mol Moles of ethanol = mass of ethanol / molar mass of ethanol Moles of ethanol = 425 g / 46.07 g/mol = 9.22 mol
02

Calculate the mole fraction of phenol

Mole fraction of phenol (X_phenol) = moles of phenol / (moles of phenol + moles of ethanol) X_phenol = 0.221 mol / (0.221 mol + 9.22 mol) = 0.0234
03

Calculate the mass percent of phenol

Mass percent of phenol = (mass of phenol / total mass of solution) * 100 Total mass of solution = mass of phenol + mass of ethanol = 20.8 g + 425 g = 445.8 g Mass percent of phenol = (20.8 g / 445.8 g) * 100 = 4.67%
04

Calculate the molality of phenol

Molality of phenol = moles of phenol / mass of ethanol (in kg) Mass of ethanol in kg = 425 g * (1 kg / 1000 g) = 0.425 kg Molality of phenol = 0.221 mol / 0.425 kg = 0.519 mol/kg The final answers are: a) Mole fraction of phenol = 0.0234 b) Mass percent of phenol = 4.67% c) Molality of phenol = 0.519 mol/kg

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