Question

1) The molarity of naphthaline in a solution of naphthalene (C10Hs, MW = 128.8), a non-ionizing and nonvolatile solute, and benzene (CH, MW = 78.11), a volatile solvent, is [CyHx) = 0.122 M. The density of the solution is D- 0.8770 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 tot [8 points) b) Xn, the mole fraction of naphthalene in the solution. [15 points)

Answer 1

1 Ca) Ciren, [colo - 0.122 M -0.122 mol/L Osmotic pressure (A) is given by the following equation T = CRT T= 0.122 mol/2x0.8821 L.atm/moth x 298.15K c= concentration = 0.122 mol/L Rogas constant - 0.0821 L.atm/matk T= 25.0°C = (25.0+2 73.15k T=298.15K T = 2.985 atm As we Know I atm = 760 ton T = 2.985 X 760 tos = 2268.4 toor π = 2268.4 tor Let Given, density of the solution (D) = 0.8770glcns moles of naphthalene (n) = = 0.122 mol Volume of saphatratene solution (V) = 1 L=1000 ml = 1000 cm3 mass of solution = D X V (mg) = 0.8770 glan3x 1000 cm3 = 877 8 mass of solution (ms) mass of naphthalene (MN) = NNX MW of Clothing = 0.122 mal x 128.8 g/mol 15.7 g mass of benzene (mp) = ms-mm = 87tg - 15-tg = 87103g moles of benzene (no) 84.88 MW of CH 78.11 g/mol mo 11.1 mol

3 Now, mole fraction of naphthalene solution (XW) กN mutno 0-122 mol 0.122 mol + 11. Imel 0.122 mol XN 0.0108 Hence, XN 11. 222. mol = 0.0108 s.

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