Given that the saturated vapour pressure of water is 3.7831*kPa at 301*K, what is the molar quantity of a 32.11*mL volume of oxygen gas collected OVER water at atmospheric pressure?

Mar 17, 2017

We can use $n = \frac{P V}{R T}$, the Ideal Gas Equation, but of course, there is a catch......

Explanation:

Water exerts an equilibrium vapour pressure, as a function of temperature. So the total pressure is the sum of this vapour pressure, and the pressure of dioxygen gas.

${P}_{\text{Gas collected"=P_"Dioxygen"+P_"SVP}}$, where ${P}_{\text{SVP"="the saturated vapour pressure}}$, which you have kindly quoted for us.

So no, we make the calculation: $1.011 \cdot a t m \equiv 1.011 \cdot \cancel{a t m} \times 101.325 \cdot k P a \cdot \cancel{a t {m}^{-} 1} = 102.4396 \cdot k P a$

And thus ${P}_{\text{Dioxygen}} = \left(102.4396 - 3.7831\right) \cdot k P a = 98.6565 \cdot k P a$

And now we plug in the numbers for the mole measurement:

${n}_{\text{dioxygen}} = \frac{98.6565 \cdot \cancel{k P a} \times 32.11 \times {10}^{-} 3 \cancel{L}}{8.314 \cancel{L \cdot k P a \cdot {K}^{-} 1} \cdot m o {l}^{-} 1 \times 301 \cdot \cancel{K}}$

$= 1.267 \times {10}^{-} 3 \cdot \frac{1}{\frac{1}{m o l}} = 1.267 \times {10}^{-} 3 \cdot m o l$

And finally, for the mass of oxygen gas, we take the product: $= 32.00 \cdot g \cdot m o {l}^{-} 1 \times 1.267 \times {10}^{-} 3 \cdot m o l \cong 40 \cdot m g$

Note that you must simply KNOW that oxygen, hydrogen, nitrogen, fluorine, and chlorine ARE diatomic molecules under standard conditions. In fact, all of the elemental gases, SAVE the Noble Gases, are binuclear.