Question

explain step by step for each
The partial molar volumes of a benzene-carbon tetrachloride solution at 25 ° C at a molar fraction of 0.5 are Vb=0.106 Lmol-1 and Vc=0.100 Lmol-1 respectively, where the subindices b and c denote C6H6 and CCl4, what is the volume of the solution formed by each of them? Since the molar volumes are C6H6=0.089 Lmol-1 and CCl4=0.097 mol-1, what is the volume change when mixing one mole of each of them? What can you deduce about the nature of intermolecular feurzas between C6H6 and CCl4?

Answer 1

Partial molar volumes of a benzene - carbon tetra- chloride solution at 250°c at a mole fraction of 0.5 are Ginen Vonzene (16)= 0.106 L mol-1 Vecea Vo = 0.100L mol a) "The wol of a solution made up of a mole of each for this we just add vb and Vc since we have only Imol of each so v= vbt Vc V = (lmoth) (0.106 Limot! b) + (1 mol c) 20.100 lmott() V- 0.206L - Thus, the volume of a solution made up of Imol of each = b) rinen, the molar volumes are C6H6 = 0.089 L mol- the molar volumes of cct = 0.097 mol-1 1. we have to find volume change when mixing one I nuole of each. 3 we do the same thing as is part (as he V = (1 mol b) co 089lmol b) + (mit () (0.097L molc) 1 0 0 0 V2 = 0.1864 This is the real volume o solution to to get the change in nolume av sv = VR-V AV = 0.186L-0206L Lov = -0.0241 (C) since av i meg.atine that means the real holume of solution is less than the theoretical volume Means molecules possess attractine forces between them to occupy less holume.