The total pressure in a flask containing air and ethanol at 25.7C is 878 mm Hg. The pressure of the air in the flask at 25.7C is 762 mm Hg. If the flask is immersed in a water bath at 450C, the total pressure is 980 mm Hg. The vapor pressure of ethanol at the new temperature is ? mm Hg.

Hint: you will need to correct the pressure of air at the new temperature using the Gas Law: P1/T1 = P2/T2

The air partial pressure increases from 762 mm Hg to (762 mm Hg) x (T2/T1)
=(762)[(273.2+450)/(273.2+25.7)= 1844 mm Hg. This is lower than the new total pressure you listed, which cannot happen. Are you sure you copied all the numbers in the problem correctly?

it's actually 45.0 C. where did u get 273.2? i replaced 450 in the equation u used, and i got 811. but that isnt correct...

Apologies for the confusion. It seems there was a mistake in the equation I provided. The correct equation to use is:

P1/T1 = P2/T2

where P1 and T1 are the initial pressure and temperature of the air in the flask (762 mm Hg and 25.7°C) and P2 and T2 are the final pressure and temperature of the air in the water bath (total pressure of 980 mm Hg and 45.0°C).

Let's plug in the values and solve for P2, the pressure of the air in the water bath:

(762 mm Hg) / (25.7°C) = P2 / (45.0°C)

Now solve for P2:

P2 = (762 mm Hg)(45.0°C) / (25.7°C)

P2 ≈ 1338.28 mm Hg

So, the pressure of the air in the water bath is approximately 1338.28 mm Hg.

To find the vapor pressure of ethanol at 45.0°C, we need to subtract the pressure of the air in the water bath from the total pressure of the flask:

Vapor pressure of ethanol = Total pressure - Pressure of air

Vapor pressure of ethanol = 980 mm Hg - 1338.28 mm Hg

Vapor pressure of ethanol ≈ -358.28 mm Hg

However, it seems there might be an error in the given data or calculations since we obtained a negative vapor pressure. Please double-check the numbers or provide additional information if possible.