# How many moles of gas exist for a volume of 0.125*L collected over water at 298*K at a pressure of 747*mm*Hg?

May 2, 2016

You need the saturated vapour pressure. This shoud have been quoted with the question.

#### Explanation:

Gas collected over water is saturated with water vapour. At $298$ $K$, the vapour pressure of water is $23.8 \cdot m m \cdot H g$.

So ${P}_{\text{Gas collected}}$ $=$ ${P}_{\text{butane") + P_("water vapour}}$

${P}_{\text{butane}}$ $=$ $\left(747 - 23.8\right) \cdot m m \cdot H g$ $=$ $723.2 \cdot m m \cdot H g$.

$n$ $=$ $\frac{P V}{R T}$ $=$ $\frac{\left(\frac{723.2 \cdot m m \cdot H g}{760 \cdot m m \cdot H g}\right) \cdot a t m \times 0.125 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 298 \cdot K}$

$\cong$ $5 \times {10}^{-} 3 \cdot m o l$.