Question

A sample of air contains 78.08% nitrogen, 20.94% oxygen, 0.0500% carbon dioxide, and 0.930% argon by volume. How many molecules of each gas are present in 6.62 L of the sample at 38°C and 2.21 atm? Enter your answers in scientific notation.

× 10 molecules N2

× 10 molecules O2

× 10 molecules CO2

× 10 molecules Ar

Answer 1

first calculate total mole of gas

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 2.21 atm,

V = volume in Liter = 6.62 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 38⁰C = 273.15+ 38 = 311.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (2.21 X 6.62)/(0.08205X 311.15) = 0.573 mole

total mole of air = 0.573 mole

calculate mole fraction of each gas

total % of air = 78.08 % + 20.94 % + 0.0500 % + 0.930 % = 100 %

mole fraction of N₂ = 78.08 / 100 = 0.7808

mole fraction of O₂ = 20.94 / 100 = 0.2094

mole fraction of CO₂ = 0.0500 / 100 = 0.0005

mole fraction of Ar = 0.930 / 100 = 0.0093

now calulate mole of each gas

mole of gas = mole fraction X total mole

mole of N₂ = 0.7808 X 0.573 = 0.4474 mole

mole of O₂ = 0.2094 X 0573 = 0.11998 mole

mole of CO₂ = 0.0005 X 0.573 = 0.0002865 mole

mole of Ar = 0.0093 X 0.573 = 0.0053289 mole

According to avogadro's law 1 mole of gas contain 6.022 X 10²³ molecule then calculate for above given mole

molecule of N₂ = 0.4474 X 6.022 X10²³ = 2.694 X 10²³

molecule of O₂ = 0.11998 X 6.022 X10²³ = 7.225X 10²²

molecule of CO₂ = 0.0002865 X 6.022 X10²³ = 1.725 X 10²⁰

molecule of Ar = 0.0053289 X 6.022 X10²³ = 3.209 X 10²¹

molecule of N₂ = 2.694 X 10²³

molecule of O₂ = 7.225X 10²²

molecule of CO₂ = 1.725 X 10²⁰

molecule of Ar =   3.209 X 10²¹