821 ML OF N2 GAS WAS COLLECTED OVER LIQUID WATER AT 300K AND 1 ATM IF VAPOUR PRESSURE IS 30 TORR THEN NO OF MOLES OF N2 GAS IN MOIST GAS MIXTURE

Answer 1

30 torr 1atm/760 torr = 0.0395 atm = vapor pressure of water

press of gas = 1 atm - 0.0395 atm = 0.96 atm

PV = nRT

n = moles = PV/RT = (0.96)(0.821)/(0.0821)(300) solve to get moles of gas