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Topic: Help: Raoult's Law problem. Finding molar mass  (Read 14626 times)

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Offline AvedisRC

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Help: Raoult's Law problem. Finding molar mass
« on: March 25, 2007, 10:54:42 PM »
I need help with a problem.  I am supposed to find the molar mass of an unkown substance and I have no idea what I have to do.  The answer I came up with was 303.03 grams.
Here's the problem:
"At a certain temperature the vapor pressure of pure benzene(C6H6) is .930 atm. A solution was prepared by dissolving 10.0g of a non dissociating, nonvolatile solute in 78.11 g of benzene (This happens to be the molar mass of benzene). The vapor pressure of the solution was found to be .900 atm. Assuming the solution behaves ideally, determine the molar mass of the solute."
So I set up:
.900 atm/.968 atm = 1 mole Benzene/ (1 mole benzene + nsolute)
How do I solve for nsolute?
Should the pressures be in torr?
Thanks
« Last Edit: March 25, 2007, 11:06:53 PM by AvedisRC »

Offline madscientist

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Re: Help: Raoult's Law problem. Finding molar mass
« Reply #1 on: March 26, 2007, 01:53:19 AM »
PoBENZENE = 0.930atm 

PBENZENE = 0.900atm

XBENZENE = PBENZENE / PoBENZENE = 0.900atm / 0.930atm = 0.968

XBENZENE = 1 mol benzene / 1 mol benzene + n mol solute

Solving for n:

n = 1 mol benzene - (0.968 * 1 mol benzene) / 0.968 =0.0331 mol

Therefore:

10g solute = 0.0331 mol solute

10g * (1/0.0331mol) = 302.1 g/mol
The only stupid question is a question not asked.

Offline AvedisRC

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Re: Help: Raoult's Law problem. Finding molar mass
« Reply #2 on: March 26, 2007, 11:17:48 AM »
Thanks, that is pretty much what I got.  I was just confused with the process at first, plus my answer of 303 seemed high because I was just thinking of salts and elements.
Thanks again.

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