An open flask sitting in a lab fridge looks empty, but it is filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?

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An open flask sitting in a lab fridge looks empty, but it is filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?

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Answer 1 · 2016-11-02T05:02:55

If we assume that the temperature inside the fridge is $5$ $5$ $^{\circ} C$ $""^@C$ , there are $0.219 \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1}$ $0.219*molxx6.022xx10^23*mol^-1$ of dinitrogen and dioxygen molecules.

Explanation:

From the Ideal Gas Law: $n = \frac{P V}{R T}$ $n=(PV)/(RT)$ $=$ $=$ $\frac{1 \cdot a t m \times 5 \cdot L}{0.0821 \cdot L \cdot a t m \cdot K^{- 1} \cdot m o l^{- 1} \times 278 \cdot K}$ $(1*atmxx5*L)/(0.0821*L*atm*K^-1*mol^-1xx278*K)$ $=$ $=$ $0.219 \cdot m o l$ $0.219*mol$

And thus we mulitply $0.219 \cdot m o l$ $0.219*mol$ by $Avogadro's number$ $"Avogadro's number"$ .

$0.219 \cdot m o l \times 6.022 \times 10^{23} \cdot m o l^{- 1} = ? ?$ $0.219*molxx6.022xx10^23*mol^-1=??$

If the flask were at room temperature, would it contain the same number of molecules?

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An open flask sitting in a lab fridge looks empty, but it is filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?

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Explanation:

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