Question

9. Calculate the freezing point of a solution made from 32.0 g octane (C8H18) dissolved in 168.0 g water. Ktof water is 5.12°C/m. 10. What is the osmotic pressure of a solution that contains 13.7 g of propyl alcohol(C3H7OH) dissolved in enough water to make 600mL of a solution at 26°C?

Answer 1

9. m = AT = Kfm m2 M, Where, m is molality of solution. X 1000 mi (multiplied by 1000, as given masses are in grams) Units of 'm' are mol/kg, To balance the equation units of 1000 are (g/kg) From above 2 equations following formula can be formulated, Kf X 1000 x m2 AT= M2 X m1 (1) Where; AT = is depression in freezing point, Kf is freezing point depression constant (= 5.12 °C/m), my is mass of SOLVENT (= 168.0 g), m2 is mass of SOLUTE (= 32.0 g), M2 is molecular mass of SOLUTE (= 114.23 g/mol).

Substitute respective values in equation (1); m °C 5.12 X 1000 X 32.0 g AT= 114.23 x 168.0 g mol °C 5.12 X 1000 X 32.0 AT=- m 114.23 g X 168.0 mol (mol) g °C AT = 0.00854 - x 1000 m #Units of Ke are °C/m. we can write it as °C/(mol/kg). Also use the units of 1000 = g/kg (as shown above in RED) AT= 0.00854 X 1000 °C mol kg (6) AT = 8.54 °C

10. TI = CRT where, C is molarity of solution. From above 2 equations following formula can be formulated, moles of solute Volume of solution (L) TT = m2 X RXT M2 X V (2) Where; Il is osmotic pressure, Tis temperature (= 26 °C = 299 K), Ris gas constant (= 0.0821 Latm/mol K) mis mass of SOLUTE (= 13.7 g), M2 is molecular mass of SOLUTE (= 60.1 g/mol), V is volume of solution (= 600 mL = 0.6 L). Substitute respective values in equation (2); 13.7 g X 0.0821 Imolatim) X 299 K T = 60.1 X 0.6 L mol = 9.33 atm

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