Question

Part 3. Problems 1) The molarity of naphthaline in a solution of naphthalene (CoHs, MW = 128.8), a non-ionizing and nonvolatile solute, and benzene (CH, MW = 78.11), a volatile solvent, is (C)•Hy] = 0.125 M. The density of the solution is D = 0.8781 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)

Answer 1

We know, & smotic pressure, CRT conc &. 125 M ot the solution - R = universal gas constant = 0.0821 L-atm.i' mot! Temperature 25'c = (25+273 ) K = 2986 (0.125* 0.0821 X298) atm 3.058 225 atm = ( 3.058 225 X 760 760) torr ? Jam= 760 torn] 23 24 25 torr (b) smotarity of naphthalene Solutions 0-125M 1000 ml Solution contain 125 moles naphthalene = (0.125 x 128-8)g naphthalene 16.1 g naphthalene NOW 1000 mL solution Solution = (1000 x 0.8781) g 878.1 g Weight of benzene - weight of solution - height of naphthalene) (878.1 - 16.1.) 8628 Mole of benzene (no) = 862 78.11 mole 11.036 mole Mole no of napthalene (un) = moke 0.25 nn Mole fraction of naphthalene nnt no 0.125 (0.125 + 11.036) 0.0112