What pressure (in torr) is exerted by 10.0 g of O2 in a 2.50 L container at a temperature of 27 degree C?

The pressure exercited by the gas will be ##”2300 torr”##.
So, you know that you have a certain amount of gas, 10.0 g to be exact, in a 2.5-L container at ##27^@”C”##. You can use the equation to solve for the pressure in atm, then use a simple conversion factor to go from atm to torr.
The number of moles of oxygen present in the container is
##”10.0″cancel(“g”) * “1 mole O”_2/(“32.0″cancel(“g”)) = “0.3125 moles O”_2##
So,
##PV = nRT => P = (nRT)/V##
##P = (“0.3125″cancel(“moles”) * 0.082(“atm” * cancel(“L”))/(cancel(“mol”) * cancel(“K”)) * (273.15 + 27)cancel(“K”))/(2.50cancel(“L”))##
##P = “3.077 atm”##
Since 1 atm is defined as being equal to 760 torr, you’ll get
##”3.077″cancel(“atm”) * “760 torr”/(cancel(“1 atm”)) = “2338.5 torr”##
Rounded to two , the number of sig figs given for 27 degrees Celsius, the answer will be
##P = color(red)(“2300 torr”)##

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