60.0 mL of O2 was collected over water at a total pressure of 755 torr and a temp of 25C. The vapor pressure of water at 25C is 24 torr. How many moles of O2 were collected?
Could you please explain and show the steps? I know the answer is 2.36x10^-3.
Ptotal = pO2 + pH2O
755 torr = pO2 + 24
pO2 = 755-24 = 731 torr
P in atm = 731/760 = ?
V = 0.060L
T = 298 K
R = 0.08206 L*atm/mol*K
Use PV = nRT and solve for n.
To find the number of moles of O2 collected, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Step 1: Convert the given total pressure to atmospheres by dividing by 760 torr (since 1 atmosphere is equal to 760 torr).
Total pressure in atmospheres = 755 torr / 760 torr/atm ≈ 0.9934 atm
Step 2: Subtract the vapor pressure of water from the total pressure to obtain the partial pressure of O2.
Partial pressure of O2 = Total pressure - Vapor pressure of water
Partial pressure of O2 = 0.9934 atm - 0.0316 atm (converted from 24 torr) = 0.9618 atm
Step 3: Convert the given volume to liters by dividing by 1000 mL/L.
Volume in liters = 60.0 mL / 1000 mL/L = 0.0600 L
Step 4: Convert the given temperature to Kelvin by adding 273.15.
Temperature in Kelvin = 25°C + 273.15 = 298.15 K
Step 5: Rearrange the ideal gas law equation to solve for the number of moles (n).
n = PV / RT
Step 6: Plug in the values into the equation:
n = (0.9618 atm * 0.0600 L) / (0.0821 L⋅atm/(mol⋅K) * 298.15 K)
Step 7: Calculate:
n ≈ 0.001536
Therefore, the number of moles of O2 collected is approximately 0.001536 moles, which is equivalent to 1.54 x 10^-3 moles (rounded to three significant figures).