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4. A sample of oxygen gas has a volume of 6.42 L at 29°C and 900.0 torr. How many oxygen molecules does it contain? (8 pts) 5
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Answer #1

Solution

PV = nRT

Where, P(Pressure) = 900.0 torr = 1.184 atm, V(Volume) = 6.42 L, n (Number of oxygen mol) = ?, R (Gas constant) = 0.082 L atm K-1 mol-1, T(Temperature) = 29°C = 302.15 K

1.184 atm × 6.42 L = n × 0.082 L atm K-1 mol-1 × 302.15 K

n = 1.184 atm × 6.42 L/0.082 L atm K-1 mol-1​​​​​ × 302.15 K

n = 7.6013 /24.776 mol = 0.306 mol

We know that

Number of molecule = Number of mol × 6.023 × 1023 molecule/mol

Number of molecule = 0.306 mol × 6.023 × 1023

Number of molecule = 1.843 × 1023 molecule

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