The volume of 02 collected over water at 25 degree C and 740 torr, was 250 mL. What volume will the dry 02 occupy at STP? (Vapor pressure of water at 25 degre C is 23.8 torr).
Use (P1V1/T1) = (P2V2/T2)
For P1 use 740 torr-23.8 = ??
Standard conditions for P2and T2.
Don't forget T is in Kelvin.
To find the volume of dry oxygen (O2) at standard temperature and pressure (STP), we need to use the ideal gas law. The ideal gas law equation is: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
At 25 degrees Celsius, the vapor pressure of water is 23.8 torr. Since the oxygen gas is collected over water, we need to subtract the vapor pressure of water from the total pressure to get the partial pressure of oxygen.
The partial pressure of oxygen (P) is calculated by subtracting the vapor pressure of water (Pwater) from the total pressure (Ptotal):
P = Ptotal - Pwater
Ptotal is given to be 740 torr, and Pwater is given to be 23.8 torr. Substituting these values, we get:
P = 740 torr - 23.8 torr = 716.2 torr
Now, at STP, the pressure is 1 atmosphere (atm) or 760 torr, and the temperature is 0 degrees Celsius or 273.15 Kelvin (K).
We can use the ratios between the volumes and the temperatures to find the volume of dry oxygen at STP.
(Volume at STP) / (Volume at 25°C) = (Temperature at 25°C) / (Temperature at STP)
(Volume at STP) / 250 mL = (25 + 273.15) K / 273.15 K
Let's calculate the volume at STP:
(Volume at STP) = 250 mL * [(25 + 273.15) K / 273.15 K]
Finally, we can plug the values into the equation to find the volume of dry oxygen at STP.