Am I doing this vapour pressure problem correctly?

[Eclair_de_XII]

Homework Statement

'The vapour pressure of ethanol (C₂H₅OH) at 19°C is 40.0 torr. A 1.00-g sample of ethanol is placed in a 2.00 L container at 19°C. If the container is closed and the ethanol is allowed to reach equilibrium with its vapour, how many grams of liquid ethanol remain?'

Homework Equations

PV = nRT
R = 62.36 L⋅torr / mol⋅K

The Attempt at a Solution

Basically, when I tried to plug in all the numbers into the ideal gas equation (with the given mol number of 1/46), it wasn't quite equal. Then I figured, that not all the ethanol is converted into vapour; only some of it. And this equation really only applies to gases (a.k.a. vapours). I didn't know how much, so that's basically what I set out to find out with what I was given.

(40.0 torr)(2.00 L) = (x mol)(62.36 L⋅torr/1 mol⋅K)(292 K)
80 L⋅torr = x19209.12 L⋅torr
x = 0.0044 mol C₂H₅OH (g)
0.0044 mol(46.06844 g/1 mol) = 0.203 g C₂H₅OH (g)
1.00 - 0.203 = 0.797 g C₂H₅OH (l)

 

[Borek]

Looks OK to me.

 

[Eclair_de_XII]

Thank you.