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 4.65 L of the sample at 49°C and 4.97 atm? Enter your answers in scientific notation.

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 atm = 4.97

V = volume in Liter = 4.65 L

n = number of mole = ?

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

T = Temperature in K = 49⁰C = 273.15+ 49 = 322.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (4.97 X 4.65) / (0.08205X 322.15) = 0.8743 mole

total mole of gas = 0.8743 mole

According to avogasro's law 1 mole of gas contain 6.022 X 10²³ molecule then 0.8743 mole of gas =

6.022 X 10²³ X 0.8743 = 5.265 X 10²³ molecule

air sample contain total 5.265 X 10²³ molecule

5.265 X 10²³ molecule = 100 % then

78.08 % nitrogen = 5.265 X 10²³ X 78.08 /100 = 4.11 X 10²³ molecule

Nitrongen molecule = 4.11 X 10²³ molecule

5.265 X 10²³ molecule = 100 % then

20.94 % oxygen = 5.265 X 10²³ X 20.94 /100 = 1.1 X 10²³ molecule

Oxygen molecule = 1.1 X 10²³ molecule

5.265 X 10²³ molecule = 100 % then

0.0500 % carbon dioxide = 5.265 X 10²³ X 0.0500 /100 = 2.63 X 10²⁰ molecule

Carbon dioxide molecule = 2.63 X 10²⁰ molecule

5.265 X 10²³ molecule = 100 % then

0.930 % Argon = 5.265 X 10²³ X 0.930 /100 = 4.896 X 10²¹ molecule

Argon molecule = 4.896 X 10²¹ molecule