A volume of 180 L contains a mixture of 0250 mole of N2 0250

A volume of 18.0 L contains a mixture of 0.250 mole of N2, 0.250  mole of O2 , and an unknown quantity of He. The temperature of the mixture is 0 degrees C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Solution

PV = nRT

=> 101325 x 18 x 10^-3 = n x 8.314 x 273

=> n = 0.804

=> moles of He = 0.804 - 0.5 = 0.304

=> grams = 0.304 x 4 = 1.21 grams