What is the Correct Formula for Calculating Vapour Pressure of Solutions?

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DottZakapa
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Homework Statement
Calculate the vapour pressure at 21.0°C of an ideal solution of water and methanol (CH3OH, P°=100 mmHg at 21.0°C) that contains 9.0 mol of water and 0,50 mol of methanol. Vapour pressure of water at 21.0°C is 18.7mmHg.
Relevant Equations
Solutions
Isn't the formula to be applied ##P= \chi_{solvent}P^o _{solvent}##?
in which ##P^o _{solvent}##= 18.7mmHg and

##\chi_{solvent}= \frac{mol_{solvent}}{mol_{solvent}+mol_{solute}}=\frac{9.0 mol}{9.0 mol+0,50 mol}##

but seems doesn't leads to the expected result
 
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Hi,

DottZakapa said:
Isn't the formula to be applied ##P= \chi_{solvent}P^o _{solvent}##?
Very unlikely: the pressure would be independent on the fraction solute ?

##\ ##
 
BvU said:
Hi,

Very unlikely: the pressure would be independent on the fraction solute ?

##\ ##

On the contrary, vapor pressure does depend on the fraction of the solute. Vapor pressure lowering is one of the four colligative properties that one has real-life experience with. For example, salt water boils at a higher temperature than pure water because salt water's vapor pressure equals 1 ATM at a higher temperature. :)

Link: https://www.chem.fsu.edu/chemlab/chm1046course/colligative.html
 
docnet said:
On the contrary, vapor pressure does depend on the fraction of the solute

You are:
1. missing the point of BvU's post
2. posting unrelated info (problem has nothing to do with colligative properties)
3. confusing OP (your post on Raoult's law was pointing in the right direction, now you are derailing)

None of that is helpful.
 
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ok, i was neglecting the methanol, basically i was considering it as non volatile solute, in that case I should not consider it, but here it must be taken into account. thanks