Colligative Properties Worksheet -.:: GEOCITIES.ws ::., Exams of Law

3) Calculate the freezing point of a solution made from 32.7 g of propane,. C3H8, dissolved in 137.0 g of benzene, C6H6. The freezing point of benzene.

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Colligative Properties Worksheet
1) What mass of water is needed to dissolve 34.8 g of copper(II) sulfate in order
to prepare a 0.521 m solution?
2) The vapor pressure of water at 20° C is 17.5 torr. What is the vapor
pressure of water over a solution containing 300. g C6H12O6 and
455 g of water?
3) Calculate the freezing point of a solution made from 32.7 g of propane,
C3H8, dissolved in 137.0 g of benzene, C6H6. The freezing point of benzene
is 5.50° C and its Kf is 5.12° C/m.
4) Calculate the boiling point of a solution made from 227 g of MgCl2 dissolved
in 700. g of water. What is the boiling point of the solution?
Kb = 0.512° C/m.
5) Calculate the concentration of nitrogen gas in a 1.00 L container exerting a
partial pressure of 572 mm Hg at room temperature. Henry’s law constant
for nitrogen at 25° C is 6.8 x 10-4 mol/L·atm.
6) A solution contains 21.6 g of a nonelectrolyte and 175 g of water. The water
freezes at -7.18° C and Kf = 1.86° C/m. Is the nonelectrolyte CH3OH or
C2H5OH?
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Colligative Properties Worksheet

  1. What mass of water is needed to dissolve 34.8 g of copper(II) sulfate in order to prepare a 0.521 m solution?

  2. The vapor pressure of water at 20° C is 17.5 torr. What is the vapor pressure of water over a solution containing 300. g C 6 H 12 O 6 and 455 g of water?

  3. Calculate the freezing point of a solution made from 32.7 g of propane, Cis 5.50° C and its K 3 H 8 , dissolved in 137.0 g of benzene, C 6 H 6. The freezing point of benzene f is 5.12° C/m.

  4. Calculate the boiling point of a solution made from 227 g of MgClin 700. g of water. What is the boiling point of the solution? 2 dissolved K (^) b = 0.512° C/m.

  5. Calculate the concentration of nitrogen gas in a 1.00 L container exerting a partial pressure of 572 mm Hg at room temperature. Henry’s law constant for nitrogen at 25° C is 6.8 x 10-4^ mol/L·atm.

  6. A solution contains 21.6 g of a nonelectrolyte and 175 g of water. The water freezes at -7.18° C and K (^) f = 1.86° C/m. Is the nonelectrolyte CH 3 OH or C 2 H 5 OH?

Solutions

  1. m 1 = 34.8 g CuSO (^4) m = 0.521 m CuSO 4 (aq) Æ Cu2+^ (aq) + SO 4 2-(aq) m = n/kg kg = n/m = (34.8 g CuSO 4 x 1 mol CuSO 4 /159.61 g CuSO 4 )/0.521 m mw = 0.418 kg H 2 O = 418 g H 2 O

  2. PA° = 17.5 torr mw = 455 g H 2 O m (^) s = 300. g C 6 H 12 O 6 C 6 H 12 O 6 (s) Æ C 6 H 12 O 6 (aq) Pw = Xw x Pw ° Xw = nw /(nw + ns ) nw = 455 g H 2 O x 1 mol H 2 O/18.02 g H 2 O = 25.2 mol H 2 O ns = 300. g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 /180.18 g C 6 H 12 O 6 = 1.67 mol C 6 H 12 O (^6) Pw = 25.2 mol/(25.2 mol + 1.67 mol) x 17.5 torr x 1 mm Hg/1 torr Pw = 16.4 mm Hg

  1. Pg = 572 mm Hg V = 1.00 L k = 6.8 x 10-4^ mol/L·atm T = 25° C Sg = kPg Sg = 6.8 x 10-4^ mol/L·atm x 572 mm Hg x 1 atm/760 mm Hg Sg = 5.1 x 10-4^ M

  2. m (^) u = 21.6 g Tf = -7.18° C mw = 175 g K (^) f = 1.86° C/m Tf = m x K (^) f x i m = Tf/K (^) f m = 7.18° C/(1.86° C/m) = 3.86 m m = n/kg n = m x kg = 3.86 m x 175 g x 1 kg/10 3 g = 0.676 mol n = m/MM MM = m/n = 21.6 g/0.676 mol = 32.0 g/mol The electrolyte is CH 3 OH.