How many moles of gas exist for a volume of $0.125 \cdot L$ $0.125L$ collected over water at $298 \cdot K$ $298K$ at a pressure of $747 \cdot m m \cdot H g$ $747mmHg$ ?

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How many moles of gas exist for a volume of $0.125 \cdot L$ $0.125L$ collected over water at $298 \cdot K$ $298K$ at a pressure of $747 \cdot m m \cdot H g$ $747mmHg$ ?

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Answer 1 · 2016-05-02T19:41:33

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 $298$ $298$ $K$ $K$ , the vapour pressure of water is $23.8 \cdot m m \cdot H g$ $23.8*mm*Hg$ .

So $P_{Gas collected}$ $P_"Gas collected"$ $=$ $=$ $P_{butane} + P_{water vapour}$ $P_("butane") + P_("water vapour")$

$P_{butane}$ $P_("butane")$ $=$ $=$ $(747 - 23.8) \cdot m m \cdot H g$ $(747-23.8)*mm*Hg$ $=$ $=$ $723.2 \cdot m m \cdot H g$ $723.2*mm*Hg$ .

$n$ $n$ $=$ $=$ $\frac{P V}{R T}$ $(PV)/(RT)$ $=$ $=$ $\frac{(\frac{723.2 \cdot m m \cdot H g}{760 \cdot m m \cdot H g}) \cdot a t m \times 0.125 \cdot L}{0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 298 \cdot K}$ $(((723.2*mm*Hg)/(760*mm*Hg))*atmxx0.125*L)/(0.0821*L*atm*K^-1*mol^-1xx298*K)$

$≅$ $~=$ $5 \times 10^{- 3} \cdot m o l$ $5xx10^-3*mol$ .

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How many moles of gas exist for a volume of $0.125 \cdot L$ $0.125L$ collected over water at $298 \cdot K$ $298K$ at a pressure of $747 \cdot m m \cdot H g$ $747mmHg$ ?

1 Answer

Explanation:

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How many moles of gas exist for a volume of 0.125⋅L0.125*L collected over water at 298⋅K298*K at a pressure of 747⋅mm⋅Hg747*mm*Hg?

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Explanation:

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How many moles of gas exist for a volume of $0.125 \cdot L$ $0.125L$ collected over water at $298 \cdot K$ $298K$ at a pressure of $747 \cdot m m \cdot H g$ $747mmHg$ ?