Ans_Ex4 - Nielsen CHM1041 Answer to Exercise Set#4 1 Why does fresh water evaporate faster than seawater under the same conditions Many ionic

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Nielsen Bhaskar/PC/1713-O/J 2/05 1 CHM1041 Answer to Exercise Set #4 1. Why does fresh water evaporate faster than seawater under the same conditions? Many ionic species are dissolved in seawater, among the most abundant are Cl -1 , SO 4 -2 , HCO 3 -1 , Na +1 , Mg +2 , K +1 , and Cz +2 . Their presence effectively decreases the concentration of the solvent, H 2 O, and consequently fewer H 2 O, and consequently fewer H 2 O molecules are available at the surface for vaporization. 2. The concentrations of two solutions (nonvolatile, nonelectrolytic solutes) are as follows: No. 1, 5.0 g of ethylene glycol, , in 100 g of water; 262 C H O No. 2, 8.0 g of glucose, , in 100 g of water 61 26 C H O Which solution has the higher (a) vapor pressure, (b) boiling point, (c) freezing point? 2 Solution No. 1, 5.0g C H O in 100 g H O 1 5.0 0.081 62.1 mole C H O gx m o l eCHO g = 0.081 0.81 0.100 mole molality m C H O Kg == 2 Solution No. 2, 8.0 g C H O in 100 g H O 22 1 8.0 0.044 180 mole C H O m o l g = 0.044 0.44 0.100 mole molality m C H O Kg (a) Since both solutions contain non-volatile solutes relative to the H 2 O solvent, the vapor pressures of the solutions are dependent upon the concentration of H 2 O. Therefore, as the solute concentration increases, the solvent concentration decreases, and the vapor pressure decreases. Solution no. 2, 044 m C 6 H 12 O 2 , should have a greater pressure than solution no. 1, 0.81 m C 2 H 6 O 2 . (b) Boiling point elevation, T = k b m. The greater the molaity, m, of the novolatile solute, the greater the boiling point elevation, and the higher the boiling point temperature. The BP of solution no. 1 (0.81 m) should be greater than the BP of solution no. 2 (0.44 m). (c) Freezing point depression, T = k f m. The greater the molality of the solute, the greater the freezing point depression. The solution with the lower solute con-centration (no. 2, 0.44 m) should have the higher freezing point (no. 1, 0.81 m).
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Nielsen Bhaskar/PC/1713-O/J 2/05 2 3. What are the freezing and boiling points of a solution prepared by dissolving 10.0 g of gluose, , in 40.0 g ? 61 26 C H O 2 HO 1 cos 10.0 0.0556 180. mole C H O moles glu e g x mole C H O g == 2 0.0665 ln 1.39 0.0400 mole C H O molality so m Kg H O freezing point soln, FP: 1.86 1.39 2.59 f TK m C m xm ∆= = ° = ° C 0.00 - 2.59 - 2.59 FP C C C ° = ° boiling point soln, BP: 0.512 1.39 0.712 b m C m = ° = ° C 100.000 0.712 100.712 B PC C + ° C ° m 4. The freezing point of an aqueous solution of a nonelectrolyte is –0.160ºC. What would be the expected normal boiling point of the same solution? 2 ln 0.000 , -0.160 , 0.160 pure so FP C FP C T C = ° = since f 0.160 0.0860 1.86 f TC mm KC m ° = ° BP soln : 0.512 0.0860 0.0440 b m C m x m = ° = ° C 100.0000 0.0440 100.0440 B C + ° C
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Nielsen Bhaskar/PC/1713-O/J 2/05 3 5 5. Arrange the following aqueous solutions in the expected order of decreasing freezing points: 42 2 0.01 m NaCl, 0.02 m MgSO , 0.01 m MgCl , 0.005 m KBr, 0.025m HCl, 0.015 m C H OH
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This note was uploaded on 12/12/2011 for the course CHM 2041 taught by Professor Staff during the Fall '09 term at Santa Fe College.

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Ans_Ex4 - Nielsen CHM1041 Answer to Exercise Set#4 1 Why does fresh water evaporate faster than seawater under the same conditions Many ionic

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