Question

4. (10 pts.) At ambient temperature and pressure, the molar volumes of pure liquids A and B are A 40 cm3/mole and 18 cm3/mole, respectively. Consider a liquid solution of A and B for which x 0.20. At that composition, the partial molar volumes of A and B are VA- 38 cm3/mole and Vs -15 cm3/mole, respectively. (a) (3 pts.) Calculate the mean molar volume, V, of the liquid solution. (b) (3 pts.) Calculate AVmix per mole of the liquid solution. (c) (4 pts.) Find the volumes of pure A and pure B that must be mixed to produce VT- 1000 cm3 of a liquid solution of composition-0.20.

Answer 1

4. Given,

The molar volume of a pure liquid A, VA = 40 cm³/mol

Partial molar volume of A, VA = 38 cm³/mol

Molar Volume of Pure B , VB = 18 cm³/mol

Partial molar volume of B, VB = 15 cm³/mol

(a) The mean molar volume of the solution = Xa*VA + XbVB

x_a = 0.20 ; x_b = 1 - 0.20 = 0.8

Th mean molar Volume of solution = XA*VA + XB*VB

= 0.2 * 38 + 0.8 * 15

= 7.6 + 12 = 19.6 cm³/mol

(b) If the partial molar volume were equal to molar volume in pure state then,

The molar Volume = XA*VA + XB*VB

= 0.2*40 + 0.8*18

= 8 + 14.4 = 22.4 cm³/mol

$\Delta$ V_mix = The molar volume of solution - The partial molar volumes of speicies

= (22.4 - 19.6) cm³/mol

= 2.8 cm³/mol

(c) Given,

V^T = 1000 cm³ with XA = 0.2

The molar volume at this concentration of A = 19.6 cm³/mol

The number of moles of solution = V^T/ molar volume of solution

= 1000 / 19.6 = 51.02 moles

Now, XA = 0.2

Thus the number of moles of A = 0.2 * 51.02 = 10.204 mol

The volume of Pure A required = number of moles of A * Molar volume of pure A

= 10.204 * 40 = 408.16 cm³

Now, x_b = 0.8

Thus the number of moles of B = 0.8 * 51.02 = 40.816 moles

The volume of Pure B required = number of moles of B * Molar volume of pure B

= 40.816 * 18 = 734.68 cm³