Question

A 1.35 L sample of dry air at 25 oC and 736 mmHg contains 0.0903 g of N2 plus other gases including O2, Ar and CO2. What is the partial pressure (in mmHg) of N2 in the air sample? Calculate the mole percent of N2 in the mixture.

Answer 1

1st calculate the total mol of gas

P = 736.0 mm Hg

= (736.0/760) atm

= 0.9684 atm

V = 1.35 L

T = 25.0 oC

= (25.0+273) K

= 298 K

find number of moles using:

P * V = n*R*T

0.9684 atm * 1.35 L = n * 0.08206 atm.L/mol.K * 298 K

n = 5.344*10^-2 mol

now calculate mol of N2

Molar mass of N2 = 28.02 g/mol

mass of N2 = 0.0903 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(0.0903 g)/(28.02 g/mol)

= 3.223*10^-3 mol

P(N2) = mol of N2 * total pressure / total mol

= (3.223*10^-3)*736 mmHg /(5.344*10^-2)

= 237 mmHg

Answer: 237 mmHg