Question

Consider a solution of 30 grams of naphthalene in 1500 kg of benzene. Pure Benzene Mol. Wt. 78.1 g/mol and o = 0.87 g/cm Th* = 353.2K AHvap = 30.8 kJ/mol Ti = 278.6K AH fus = 10.6 kJ/mol a) Calculate the boiling point of the solution b) Calculate the freezing point of the solution c) Calculate the osmotic pressure of the solution at 400K d) Calculate the vapor pressure of benzene in the solution at 400K e) Cryoscopic constant of benzene f) Ebullioscopic constant of benzene

Answer 1

& 30gm of napthalene in 1500 kg of benzene. So, w, 1500kg Mi = 128. 17 g/mol W2 = 30gm M₂ = 78.1 g/mol Given data, To = 353.2K, Athap = Lv = 30 8K / med T* - 278.6K Alfus=4 = 10.6 K5/mol a ) 4 a we know AH ATG Le W2 MI 4Tb tur L R = M wi To ~ 353.2- To W2 x Mix R on Tor = My Wy x l Where To Tb. To 353.2- To 308 128-178/mo 8.3145 mork! 353.24 Tb - 78.13 mot 1500x 1000g 30.8x1000 Imo => 353.2-T6 = 8.859x10-9x353.2xTb → [1+8.859 X10-9x 353-2)] To = 353.2 » T6 = 135 : Boiling point of the solution = 353.198 K 353.2 1.00000319 = 353198 - 8159. 278.6 - Tt 278.6XTJ 2 It - Tt Lt. We mi = M2 wi Ty t' R 30x128.17 X 8.314 78.4 x 1500x1000 X 10.6 X1000 7 278.6-T6 = 2.5743x10-8X278.6XTf > [+G. 1721x10-6)] It = 278.6 TH=279 € 12x1097 278 : Freezing point of the solution is 278.598K

) Density of the som. = 0.87 g/cm3 .: Osmotic pressure, IT = RT X 2 as density is given, Vi - P RTP & T = Mi x2 - RTP W2 Mi - M X M X wi 82 Caten mot lyx 400KX0.878.4 x 308x Mi Mix 78.18.mott x 1500x1000% 82 x 400 x 0.87X30 atm. 78.1 x 1500 x 1000 = 0.7307 atm Osmotic pressure of the soin is 0.7307 atm. a. d) It vapour pressure of pure benzene in Pio &' benzene in solo is P. then relative lowering of vapour pressure, => AP X2 Now, IT = P x2 = RIP ( 48 ) RT P 0.7307 atm x 128:17g. mol-1 - 8246 mot ka atm x 400k x 0.87 god. = 3.281x10 3 Pi-P, 2 3.281 X103 Pi => 1 - 2 - 3.281 X10-3 27 P 2 PO CI+ ( 3.281x103)] . It Pio is known, we can calculate p.

e) Cryoscopic RTV Tex 1000 Constant = kg = Now 4TJ = K4 x m >> (T, Tt) = K4 x W 2 > Kf = (TTf) x M2 KW, 1. Wzx1000 2786 - 278.598) X 78.1 x S 1500 X 1000 30 x1000 = 7.818. g.mot! . (ryoscopic Constant is 7.81 kg molt 1 Ebullioscopic constant = ko = eWx1000 To 4Tb kb x m. > Tbk - Tb) = Kbx W2 x 1000 M2xWI Y a Kb = (T6²-T6 x W₂ x 1000 M2 x wi W, W2X1000 1 = (353.2- 353.1987 *78.1x 1500x1000 30xiooo 2 7.81 . Ebullioscopic constant ook is also 7.81 kg moll