Question

A student finds that a mixture of acetone and water at 740 mmHg boils at 68 Celcius. Pure acetone has a vapor pressure of 1200 mmHg and water has a vapor pressure of 237 mmHg at the given temperature.

a.) What are the mole fractions of each component in the mixture?

b.) If there are 0.500 moles in the solution, what is the composition by mass, of the mixture?

Answer 1

Solution :-

Total pressure of the solution is the sum of the partial pressures of the components of the mixture. therefore using the pure vapor pressure and total pressure of the mixture lets calculate the mole fraction of the each component.

Lets calculate the mole fractions of the each

P tot = ( p acetone * molefraction of acetone)+ (p water *(1-mole fraction of water )

740 mmHg = (1200 * mole fraction of acetone ) +(237 mmHg * (1-mole fraction of water))

740 = (1200*x)+(237*(1-x))

Solving for x we get

X= 0.5223

Therefore the mole fraction of the acetone = 0.5223

And mole fraction of the water = 1-x = 1-0.5223 = 0.4777

b) If there are 0.500 moles in the solution, what is the composition by mass, of the mixture?

Using the mole fractions lets calculate the moles of the acetone and water

Moles of acetone = mole fraction * total moles

                               = 0.5223* 0.500 mol

                               = 0.2612 mol acetone

Moles of water = 0.500 mol – 0.2612 mol = 0.2388 mol

Now lets calculate the mass of the each

Mass of acetone = moles * molar mass

                              = 0.2612 mol * 58.08 g per mol

                              = 15.17 g acetone

Mass of water = moles * molar mass

                          = 0.2388 mol * 18.0148 g per mol

                          = 4.30 g