The volume of oxygen, collected over water, is 185 mL at 25 degrees Celsius and 600 torr. calculate the dry volume of the oxygen at STP

(for the question, you have to convert from mL to L, Celcius to Kelvin, and torr to atmospheres)

0.1593 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

P₁V₁T₂ = P₂V₂T₁

P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.

P₂ (at STP) = 1.0 atm, V₂ = ? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.

∴ V₂ = P₁V₁T₂/P₂T₁ = (0.789 atm) (0.185 mL) (298.0 K) / (1.0 atm) (273.0 K) = 0.1593 L.