a) partial pressure of O2 = total pressure - vapor pressure of H2O
= 752 - 19.8 =732.2 mmHg
b) moles of dry gas are calculated using ideal gas equation PV = nRT
n = PV/RT
P = 732.2 mm = 732.2/760 atm
V = 125mL = 0.125L
T= 22C = 22+273 = 295 K
Thus n = [732.2/760]atmx 0.125L /0.0821l.atm/mol.K x295K
=0.00497 mol
c) moles of wet gas calculated using ideal gas equation similarly using P = 752mm
thus
n = [752/760]atmx 0.125L /0.0821l.atm/mol.K x295K
=0.00510mol
d) 0.0250g of N2 added
moles of N2 = 0.0250g/28g/mol =8.92x10-4 mol
partial pressure of N2 = nRT/V
= 8.92x10-4 molx0.0821 L.atm/mol.K x295K/ 0.125 L
=0.173 atm =131.48 mmHg
e)Total pressure after N2 is added =[ Partial pressure of O2 + vapor pressure of H2O] + partial presure of N2
= pressure before additon + partial pressure of N2
= 752 + 131.48 =883.48 mmHg
28. A sample of oxygen is collected over water at 22°C and 752 mmHg in a...
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