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Topic: Vapor Pressure  (Read 3892 times)

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Offline Zhadows

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Vapor Pressure
« on: February 27, 2008, 07:38:55 PM »
Hi, I am just curious if someone could check an answer for me:

A) What is the vapor pressure of a solution prepared by dissolving 20.50 g of sodium nitrate in 100.0 g of water at 25 C. The vp of water at 25 C is 23.8 mm Hg.

I calculated the moles of H20 and NaNO3, did the mol fractionation of H2O [5.55] and divided it by 5.55+mol of NaNO3 * 5 because there are 5 units [1xNA+1xN+3xO], then I utilized Raoult's law which is: Psoln=Psolv*Psolv and got a number of 19.56, does this seem right?



Also, would anyone be able to provide me with some advice in solving this one?

B) What is the vapor pressure of a mixture of benzene and toluene is at 25 C in which 1/3 of the molecules are benzene, and what is the mole fraction of each component in the vapor phase? At 25 C, the vp for benzene = 96.0 mm Hg and 30.3 mm Hg for toluene.

I was thinking of approaching this the same way as I did to part A but taking the delta of both values and having that as my answer, but the 1/3 molecules just confused me, any suggestions?

Thank you.

Offline LQ43

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Re: Vapor Pressure
« Reply #1 on: February 28, 2008, 09:01:17 PM »
NaNO3 * 5 because there are 5 units [1xNA+1xN+3xO],

NaNO3 does not dissociate into 5 units in water -

what are the ions that make up NaNO3?

B) What is the vapor pressure of a mixture of benzene and toluene is at 25 C in which 1/3 of the molecules are benzene, and what is the mole fraction of each component in the vapor phase? At 25 C, the vp for benzene = 96.0 mm Hg and 30.3 mm Hg for toluene.


Ptotal = XaPoA  +  XbPob

Xa , Xb is the mole fraction or (molecule fraction...)


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