Question

Name --------- Show all work if you want to get partial credit. Good Luck! 1a. For each of the following compounds, predict whether you would expect it to be more soluble in water or hexane (C6H14). (a) 12 ------- (b) CaCl2 ---------- (c) HBr -- (d) C&H,OC4H, --- (e) CH3NH2 -------- 1b. Arrange the following aqueous solutions according to increasing freezing point: 0.2m glucose, 0.1m CaCl2, 0.1m MgSO4, 0.1m Na3PO4 (m is molality). Lowest f.p. ----- --------- highest f.p. 2. Ethylene glycol (a non-volatile compound, MW = 62) was added to 4 liters of water at 40°C and the vapor pressure of the water was found to be 40 torr. The vapor pressure of pure water at 40°C is 55 torr, the density of water is 1.00g/mL, and the density of ethylene glycol is 1.11g/mL. (a) what is the mole fraction of the water? ------- (b) how many moles of ethylene glycol were added? - (C) how many grams of ethylene glycol were added? - (c) how many liters of ethylene glycol were added? ---

Answer 1

Water polar in nature Hexane = Non polao in nature I since Iz is non polas in nature, it soluble in Hexane call, since it is polar, it is soluble in Hon (like dissolves like, HB r = exist as Ht Bro soluble in water Синя ОСчня 20otble in , ходе. it in non polar CH₂ NH & soleble in rexane.

D oem glucose olm cach olm Mgsor olm Na, Poa vonth off factor is for Non polad 21 (covalent solutes = por ionic solwe it con tave vallle = = call cat 2ct zo totale 3 ions i=3 Masoo get so 2 = totalions o i=2 Na₂ 10 4 2 3 Not Pot = total alon a i=4. freezing point olfa ikammolality of a ixm towering in since m' is same i is deciding factor NO 3 PO > Cach, Mgida - - more is valell ofi, more is depression in freezing point ut, depressin in freeing point increases. low is vallue of TF. more - Napo, have lowest freezing point — No, PD, Ca, MgO, , шосе llowe Fly rigne F.P) nen)

- - 0B Relative lowering in vapour pressure = PA-PS PS NA PAC pure water vapou ou preanures a 55toor. Po l solurion vapour prennures zaotoor. nB = molesof solute - nA = moles of solvent. Sementara - 48103 40 Чxi o3 NB = 15 x 4x103 40 YIB n8= 33- 33 moles of ethylene glycol 6 mole, ef ethylene gly ell a weight M.W (62 weight = 83.33x62 weight = 5166.66g / density = man volume. Tills mart volunre: 5166.66 volume = 4654.65 me - 4652