Oxygen gas generated in an experiment is collected at 25.0°c in a bottle inverted in a trough of water. the total pressure is 1.000 atm corrected for atmospheric pressure. when the water level in the originally full bottle has fallen to the level in the trough, the volume of collected gas is 1750 ml. how many moles of oxygen gas have been collected given the vapor pressure of water at 25.0°c is 23.8 mm hg?

0.007 mol

Explanation:

We can solve this problem using the ideal gas law:

PV = nRT

where P is the total pressure, V is the volume, R the gas constant, T is the temperature and n is the number of moles we are seeking.

Keep in mind that when  we collect a gas over water we have to correct for the vapor pressure of water at  the temperature in the experiment.

Ptotal = PH₂O + PO₂  ⇒ PO₂ = Ptotal - PH₂O

Since R constant has unit of Latm/Kmol we have to convert to the proper unit the volume and temperature.

P H₂O = 23.8 mmHg x 1 atm/760 mmHg =  0.031 atm

V = 1750 mL x 1 L/ 1000 mL = 0.175 L

T = (25 + 273) K = 298 K

PO₂ = 1 atm - 0.031 atm = 0.969 atm

n =  PV/RT = 0.969 atm x  0.1750 L / (0.08205 Latm/Kmol x 298 K)

n = 0.007 mol

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there are no statements provided but here is why:

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