Question

urgent

1) The molarity of naphthaline in a solution of naphthalene (C1H, MW = 128.8), a non-ionizing and nonvolatile solute, and benzene (CH, MW = 78.11), a volatile solvent, is [CroHs) = 0.086 M. The density of the solution is D- 0.8748 g/cm'. Find the following: a) The osmotic pressure exerted by the solution, at T = 25.0 °C. Give your final answer in units of torr. [8 points) b) Xn, the mole fraction of naphthalene in the solution. [15 points) I

Answer 1

π = i CRT where i = Vant's Hoff factor = 1 (non-ionizing dolute) Molarity of Aolution = 0,086 M RA 0.082) Lash l kmol T temp = 298 K -25° c = C = TT Molarity = no. of moles Peu unit rolume 0.086 M solution osmotic Pressure el) (01086) (0.0621) (998) atm 2.1041 atm π = T 760 torr 760 x 2,1041 =1599.116 torr XI 1 atom- Doloul atm moles of napthalene Mole fraction of natthelene meles of hap thalenet nales of Benzene density of solution = 0.8748 g/ m 3 = 0.87469) me 0.87488 874.8 g/L Vioool molarity of xolution - moles of napthalene Volume of solmin Oo86 M je 0.1086 moles of nap the lene in 1 L of Bol" t solution volume density of solchen = 874,8 g/L mom of solchon volume x density = 874189 1 L - 2

nom man of solution = 874.8g moles of napthalene = 0.086 many nabthalene in sol", moles x molecular of napthelone 01066 x 128.8 11.07689 man of benzene man of Allahon - man of nap the bene = 874,8 - 11.0768 = g 863,72329 man of Benzene moles of Benzene = molar mom of Benzene 8 6317239 78011 = 1.0578 males mole fraction of napthelene a 0,086 01086 +1.0578 CioB6 0.1438 -0.00772 bolo = 0.00772