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Topic: Solutions - Conversions  (Read 7624 times)

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Offline MitchTwitchita

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Solutions - Conversions
« on: March 08, 2008, 04:42:09 PM »
Fill in the blanks in the table for aqueous solutions of the compounds shown.  The density of the solution is in g/mL.  Abbreviations: m is molality, Mass % is the mass percentage of the solute, Mole Frac. is the mole fraction of the solute, and M is the molarity of the solute.

Comound     f.w.         d.soln.         m               Mass%          Mole Frac.           M

C3H6O         58.05      0.9874                           8.000               
H3PO4         98.00      1.1804       4.373
NaCl            58.44      1.0894                                                                    2.330


For C3H6O,

Molality: Assume out of 100 g,

8g C3H6O x (1 mol C3H6O/58.05g C3H6O) = 0.1378 mol.
0.1378 mol/0.09200 kg = 1.498 mol/kg

Mole Fraction:

0.1378/0.1378 + (92.00g H2O x 1 mol H2O/18.02g H2O)
=0.1378/(0.1378 + 5.105)
=0.02628

Molarity:

(1.498 mol C3H6O x 58.05g C3H6O/1 mol C3H6O) + 1000g H2O = 1087g
V = m/d = 1087g/0.9874g/mL = 1101 mL x 1L/1000 mL = 1.101 L
M = n/V = 1.498 mol/1.101 L = 1.361 mol/L


For H3PO4,

Molarity:

(4.373 mol H3PO4 x 98.00g H3PO4/1 mol H3PO4) + 1000g H2O = 1429 g solution
V = m/d = (1429g/1.1804 g/mol) x (1 L/1000 mL) = 1.211 L
M= n/V = 4.373 mol/1.211 L =  3.611 mol/L

Mass %:

Mass of solute/ mass of solution
=(428.6 g/1429g) x 100%
=29.99%

Mole Fraction: Assume out of 100g,

29.99 g H3PO4 x (1 mol H3PO4/ 98.00 g H3PO4) = 0.3060 mol H3PO4
70.01 g H2O x (1 mol H2O/ 18.02 g H2O) = 3.885 mol H2O
0.3060/(0.3060 + 3.885)
=7.301 x 10^-2


For NaCl,

Molality:

1 L soln x (1000 mL soln/1 l soln) x (1.0894g/1 mL soln) = 1089
mass H2O = mass solution - mass solute
=1089 g - (2.330 g NaCl x 58.44 g NaCl/ 1 mol NaCl)
=952.8
molality = 2.330 mol/0.9528 kg
=2.445 mol/kg

Mass %:

mass solute/mass solution
=(136.2 g/1089g) x 100%
=12.51%

Mole Fraction:  Assume out of 100g,

12.51 g NaCl x (1 mol NaCl/58.44 g NaCl) = 0.2141 mol NaCl
87.49 g H2O x (1 mol H2O/18.02 g H2O) = 4.855 mol H2O
0.2141/(0.2141 + 4.855)
=4.224 x 10^-2

Does this look right to anybody?  If not, could you please let me know where I'm going wrong.

Offline Borek

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Re: Solutions - Conversions
« Reply #1 on: March 08, 2008, 04:58:41 PM »
What do you mean by C3H6O?
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Offline MitchTwitchita

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Re: Solutions - Conversions
« Reply #2 on: March 08, 2008, 05:27:36 PM »
Thanks Borek!  I don't know what they mean by C3H6O, that's just the formula I was given.  Would it by cyclo-propanol?

Offline Borek

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Re: Solutions - Conversions
« Reply #3 on: March 08, 2008, 05:31:12 PM »
Could be acetone as well. But in fact - if density is given - it doesn't matter.

Note that density tables built into CASC slightly differ from the data you were given, so some (if not most) of the results are slightly off. But they are close enough to confirm (or deny) your calculations. Honestly - I have not checked them thoroughly.
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Offline MitchTwitchita

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Re: Solutions - Conversions
« Reply #4 on: March 08, 2008, 05:50:16 PM »
Thanks Borek, they look pretty good to me!!!!

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