Molar Mass of Unknown Solute and Mole Fraction of Gases


by Soaring Crane
Tags: fraction, gases, mass, molar, mole, solute, unknown
Soaring Crane
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#1
Jan27-06, 10:10 AM
P: 483
1) Find the molar mass of an unknown substance if 227.5 mL of an aqueous solution containing 2.785 grams of the unknown generates an osmotic pressure of 588.0 mm Hg at 20.9 Celsius. Report your answer to 3 significant figures.

pi = MRT

pi = 588.0 mm Hg = 588 mm Hg/760 mm Hg = 0.7736842 atm
T = 20.9 C = 20.9 C + 273.15 = 294.05 K

molar mass = g solute / mol solute

pi = R*T*(mol solute/L solution)

mol solute = (pi*Volume_solution)/(R*T) = (0.7736842 atm * 0.2275 L)/(294.05 K * 0.08206) = 0.007294448 mol

m. mass = 2.785 g / 0.007294448 mol = 381.797 g/mol = 382 g/mol ?????


2) Calculate the mole fraction of benzene in the vapor phase from a solution that contains a mole fraction of benzene of 0.557 when mixed with toluene if the vapor pressure of pure benzene is 183 mmHg and that of toluene is 59.2 mm Hg.

X_b = 0.557-------------------VP_b = 183 mm Hg
X_t = 1 - 0.557 = 0.443-------VP_t = 59. 2 mm Hg

Total (vapor) pressure = (X_b*P_b) + (X_t*P_t)
P_t = 0.557*183 mm Hg + 0.443*59.2 mm Hg
P_t = 101.931 mm Hg + 26.2256 mm Hg = 128.1566 mm Hg

X_gas = P_gas (benzene) / P_total = 101.931 mm Hg / 128.1566 mm Hg = 0.795 ????


Thank you.
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Soaring Crane
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#2
Jan30-06, 08:44 PM
P: 483
Can anyone please help me?
Astronuc
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#3
Jan30-06, 08:50 PM
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I think the approach to a) is correct, but let me get GCT or Ouabache to take a look.

Plastic Photon
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#4
Jan31-06, 01:49 PM
P: 166

Molar Mass of Unknown Solute and Mole Fraction of Gases


#1 is correct
Soaring Crane
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#5
Feb1-06, 10:24 PM
P: 483
Is #2 correct?

Thanks again.
aalmighty
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#6
Feb1-06, 10:45 PM
P: 17
#2 is correct too.
GCT
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#7
Feb2-06, 02:51 PM
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yeah, both of them seem fine.
Ouabache
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#8
Feb2-06, 07:35 PM
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Nice job Soaring Crane.. I also tried them and come up with those solutions.


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