I have a cotton ball inside a 35.8 cm3 cylinder, at 25C normal pressure. If I add 0.5ml of 50% ethanol to the cotton ball and quickly seal the cylinder, what is the internal vapor concentration over time?

How much pure ethanol? What is the vapor pressure of Ethanol? "internal vapor concentration over time" is a long way to say equilibrium. Remember, this is a sealed tube so nothing evaporates.

You have the volume of the cylinder and temperature inside the cylinder. Calculate the moles of ethanol and use Ideal gas Equation.For more help show your attempt first.

You have the volume of the cylider and temperature inside it. You just need to calculate the moles of ethanol and use Ideal gas equation.
To get more help you need to show your solution first.

Thank you for the comment! Here's what I've got, but it still doesn't make sense to me:
Ethanol MM = 46.06844 g/mol
Water MM = 18.0148 g/mol
Temperature = 25C or 298.15K
Volume = 0.358 m3

With 0.25 ml of water and 0.25 ml ethanol, I have 0.0139 moles of water and 0.00543 moles of ethanol

PV = nRTP * 0.358 m3 = 0.00543 mol ethanol * 8.314 J/K mol * 298.15K
P = 37.7 pa

I still don't know what this means about the internal vapor concentration over time. Is the vapor 5% ethanol after 5min, 20min an 1hr? And if left to reach equilibrium, what is the highest concentration of ethanol vapor reached? And what about using Raoult's Law to find vapor pressure of an ethanol + water mixture... like (0.00553) / (0.0139 + 0.0053) *? Torr

See 50% solution means 50% of the mass of the solution is ethanol.
Now, you get the density of the solution and then use it to find the mass & hence moles of ethanol.