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
4. A sample of oxygen gas has a volume of 6.42 L at 29°C and 900.0...
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