How many moles of air are in 1.00 L at - 23 C anf 101 KPa?

0.041 mol.

Explanation:

To solve this problem, we can use the general law of ideal gases:

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.