The densities of pure water and ethanol are 997 and 789 kg m-3, respectively. For xethanol = 0.35, the partial molar volumes of ethanol and water are 55.2 and 17.8*10-3 L mol-1, respectively. Part ACalculate the change in volume relative to the pure components when 2.50 L of a solution with xethanol = 0.35 is prepared. Express your answer to two significant figures and include the appropriate units.

the deviation for ethanol relative to the pure components is - 0.09 L (contracts) and for water is 0.01 L (expands)

Explanation:

Assuming we are preparing 2.50 L solution of ethanol-water without considering the change in volume

x et / 1 - x et = n et / n w = (M w/M et) * m et / m w → m et = (x et / 1 - x et) * k * n w

V final = m w / D w + m et / D et = m w / D w + (x et / 1 - x et) * k / D et * m w

therefore

m w = V final / (1/D w + [ (x et / 1 - x et) * k / D et)

replacing values

m w = 2.50 L / (1 / 997 kg/m³ + (0.35/0.65) * (46/18) / 789 kg/m³) * m³ / 1000 L = 0.91 Kg

m et = (0.35/0.65) * (46/18) 0.91 Kg = 0.49 Kg

therefore

V w = m w / D w = 0.91 Kg / 997 Kg/m³ * 1000 L/m³ = 0.91 L

V et = m w / D w = 0.49 Kg / 789 Kg/m³ * 1000 L/m³ = 1.59 L

n w = 0.91 Kg / 18 g/mol * 1000 g/Kg = 50.55 moles of water

n et = (0.35 / 0.65) * 50.55 = 27.22 moles of ethanol

the real volumes in solution are

Vr et = n et * v et = 27.22 mol * 55.2*10^-3 L/mol = 1.50 L

Vr w = n et * v et = 50.55 mol * 17.8*10^-3 L/mol = 0.90 L

the deviation relative to pure components are

d et = Vr et - V et = 1.50 L - 1.59 L = (-0.09 L)

d w = Vr w - V w = 0.91 L - 0.90 L = 0.01 L