Chemistry: A sample of air contains 78.08% nitrogen, 20.94% oxygen, 0.0500% carbon dioxide, and 0.930% argon by volume. How many molecules of each gas are present in 6.51 L of the sample at 43°C and 1.59 atm? Enter your answers in scientific notation.

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A sample of air contains 78.08% nitrogen, 20.94%

Ответ:

zoieihazel

24.01.2022

Chemistry

Answer : The number of molecules of nitrogen, oxygen, carbon dioxide and argon is, $1.88\times 10^{23}$ , $5.03\times 10^{23}$ , $1.19\times 10^{20}$ and $2.23\times 10^{21}$ respectively.

Explanation :

First we have to calculate the total moles of mixture of gas by using ideal gas equation.

PV=nRT

where,

P = Pressure of mixture of gas = 1.59 atm

V = Volume of mixture of gas = 6.51 L

n = number of moles mixture of gas = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of mixture of gas = 43^oC=273+43=316K

Putting values in above equation, we get:

$1.59atm\times 6.51L=n\times (0.0821L.atm/mol.K)\times 316K$

n=0.399mol

Now we have to calculate the moles of nitrogen, oxygen, carbon dioxide and argon.

Moles of nitrogen = $78.08\% \times 0.399=\frac{78.08}{100}\times 0.399=0.312mol$

Moles of oxygen = $20.94\% \times 0.399=\frac{20.94}{100}\times 0.399=0.0836mol$

Moles of carbon dioxide = $0.0500\% \times 0.399=\frac{0.0500}{100}\times 0.399=0.000199mol$

Moles of argon = $0.930\% \times 0.399=\frac{0.930}{100}\times 0.399=0.00371mol$

Now we have to calculate the number of molecules of nitrogen, oxygen, carbon dioxide and argon.

As, 1 mole of nitrogen contains $6.022\times 10^{23}$ number of molecules of nitrogen.

So, 0.312 mole of nitrogen contains $0.312\times 6.022\times 10^{23}=1.88\times 10^{23}$ number of molecules of nitrogen.

and,

As, 1 mole of oxygen contains $6.022\times 10^{23}$ number of molecules of oxygen.

So, 0.0836 mole of oxygen contains $0.0836\times 6.022\times 10^{23}=5.03\times 10^{23}$ number of molecules of oxygen.

and,

As, 1 mole of carbon dioxide contains $6.022\times 10^{23}$ number of molecules of carbon dioxide.

So, 0.000199 mole of carbon dioxide contains $0.000199\times 6.022\times 10^{23}=1.19\times 10^{20}$ number of molecules of carbon dioxide.

and,

As, 1 mole of argon contains $6.022\times 10^{23}$ number of molecules of argon.

So, 0.00371 mole of argon contains $0.00371\times 6.022\times 10^{23}=2.23\times 10^{21}$ number of molecules of argon.

Therefore, the number of molecules of nitrogen, oxygen, carbon dioxide and argon is, $1.88\times 10^{23}$ , $5.03\times 10^{23}$ , $1.19\times 10^{20}$ and $2.23\times 10^{21}$ respectively.

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A sample of air contains 78.08% nitrogen, 20.94% oxygen, 0.0500% carbon dioxide, and 0.930% argon by volume. How many molecules of each gas are present in 6.51 L of the sample at 43°C and 1.59 atm? Enter your answers in scientific notation.

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