Question

An apartment has a living room whose dimensions are 2.5 m 4.8 m 5.9 m. Assume that the air in the room is composed of 79% nitrogen (N2) and 21% oxygen (O2). At a temperature of 27°C and a pressure of 1.01 105 Pa, what is the mass (in grams) of the air?
g

Answer 1

V = 2.5 * 4.8 * 5.9 = 70.8 m^3

N is 14 so N2 is 28 grams/mole
O is 16 so O2 is 32 grams/mole
so a mole of "air" has mass of .79*28 + .21*32 = 28.8 grams/mole

assume air is ideal gas
P V = n R T
1.01 * 10^5 * 70.8 = n * 8.3145 *(273+27)
n = 3037 moles
3037* 28.8 = 87468 grams

Answer 2

my teacher says this is the incorrect answer. what could be wrong