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”)##