How many moles of air are in 1.00 l at – 23 c anf 101 kpa?

0.041 mol.

Explanation:

PV = nRT,

where, P is the pressure of the air = 101 kPa = 0.9967 atm.

V is the volume of the air = 1.00 L.

n is the no. of moles of air = ??? mol.

R is the general gas constant = 0.082 L.atm/mol.K.

T is the temperature of the air = 23 °C + 273 = 296.0 K.

∴ n (no. of moles of air) = PV/RT = (0.9967 atm)(1.00 L)/(0.082 L.atm/mol.K)(296.0 K) = 0.041 mol.