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

Is it not just this ?

Explanation: