Question

I will rate, please answer.

acetone and toulene are both volatile liquids. At 25 C , the vapor pressure of pure acetone and pure otulene are 35.2 torr and 77.9 torr, respectively.

a) Sketch a Raoult's Law plot for this 2 component solution, assuming ideal behavior, and calculate the vapor pressure above a solution containing 0.767 mole of acetone and 0.553 mole of toulene

b) The vapor pressure of the above oslution (in a) was experimentally measured and was found tob e 62.8 torr. Exokain the significance of the experimental results and explain whether the sign of the delta H of the solution is + or -.

Answer 1

Raoult's law : At any given temperature the partial vapour pressure of any component of a solution is equal to the product of mole fraction and vapour pressure of this component in the pure state

Partial vapour pressure = mole fraction x vapour preessure of pure solvent

Let Pa, x_a, p_{a , Na} and Pb,x_b, p_b N_bbe the parital pressure, mole fraction, vapour pressure in pure state, number of moles of Acetone and toluene respectively

given, Na = 0.767, Nb = 0.553

So, Total mole (N) = Na+Nb 0.767+0.553 = 1.32

Hence mole fractions x_a = Na/N 0.767/1.32 = 0.5811

and x_b = Na/N = 0.553/1.32 = 0.4189

Now, from roult law we can write expression for partial vapour pressures as,

Pa = x_a x p_a and Pb = x_b x P_b.

Pa = 0.5811 x 35.2 and Pb = 0.4189 x 77.9

Pa = 20.46 torr and Pb = 32.63 torr

Then total vapour pressure of solution of Acetone+toluene = P = Pa+ Pb = 20.46+32.63 = 53.09 torr.

Calculate (Raoult's law predicted) vapour pressure of solution = 53.09 torr

Ans experimentally calculated (observed) vapour pressure of solution = 62.8 torr

i.e. Obsereved vapour pressure is greater than that of predicted by Raoult's law. This indicate that the new interaction in solution are weaker than in the individual pure solvent and in this situation Delta H (Enthalpy change of mixing) will be Poitive (+Ve).