2. A eudiometer contains 45 ml of oxygen when the barometer reading is 731.7 mm and the temperature is 27* C . The water level inside the tube is 68mm higher than that outside . What volume does the dry oxygen occupy at S.T.P?

It is not impossible.

Find the pressure of the O2 collected, you know the volume.

Pressure=731.7-watervaporpressure.

You get the water vapor pressure from a chart. http://home.comcast.net/~frankhanson2/vapor.htm

Now, knowing the pressure, temp, and volume of the O2, convert it to volume at StP

P1V1/T1=P2V2/T2 solve for V2

To find the volume of dry oxygen at Standard Temperature and Pressure (STP), we need to use the Ideal Gas Law equation:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal Gas Constant
T = Temperature

Given information:
Initial volume of oxygen in the eudiometer (V1) = 45 ml = 45 cm³
Barometer reading (Pressure outside) (P1) = 731.7 mm
Temperature (T1) = 27°C = 273 + 27 K = 300 K
Water level difference (Δh) = 68 mm

First, we need to convert the volume from ml to cm³:
45 ml * 1 cm³/ml = 45 cm³

Then, we need to adjust the barometer reading to account for the water level difference.

Pressure inside the eudiometer (P2) = P1 + Δh
P2 = 731.7 mm + 68 mm = 799.7 mm

Next, we need to convert the pressure from mmHg to atm (since R is given in atm):
799.7 mmHg * 1 atm/760 mmHg = 1.052 atm

Now we can calculate the number of moles in the eudiometer using the Ideal Gas Law:

PV = nRT
n = PV / RT

n = (1.052 atm) * (45 cm³) / (0.0821 atm·L/mol·K) * (300 K)
n = 0.0537 mol (rounded to 4 decimal places)

Since 22.4 L of any gas at STP contains 1 mole of gas, we can now find the volume of dry oxygen at STP:

V2 = n * 22.4 L/mol
V2 = 0.0537 mol * 22.4 L/mol
V2 = 1.202 L (rounded to 3 decimal places)

Therefore, the volume of dry oxygen at STP is approximately 1.202 liters.