Question

How heavy is 1.00 L of each of the following gases at 1 atm pressure and 25.00 °C: a. O2 b. N2 c. CO₂ d. H2O

Answer 1

T = 25.00 oC

= (25.00 +273.15) K

= 298.15 K

a)

Given:

P = 1.0 atm

V = 1.0 L

T = 298.15 K

find number of moles using:

P * V = n*R*T

1 atm * 1 L = n * 0.08206 atm.L/mol.K * 298.15 K

n = 4.087*10^-2 mol

Molar mass of O2 = 32 g/mol

use:

mass of O2,

m = number of mol * molar mass

= 4.087*10^-2 mol * 32 g/mol

= 1.308 g

Answer: 1.31 g

b)

Given:

P = 1.0 atm

V = 1.0 L

T = 298.15 K

find number of moles using:

P * V = n*R*T

1 atm * 1 L = n * 0.08206 atm.L/mol.K * 298.15 K

n = 4.087*10^-2 mol

Molar mass of N2 = 28.02 g/mol

use:

mass of N2,

m = number of mol * molar mass

= 4.087*10^-2 mol * 28.02 g/mol

= 1.145 g

Answer: 1.14 g

c)

Given:

P = 1.0 atm

V = 1.0 L

T = 298.15 K

find number of moles using:

P * V = n*R*T

1 atm * 1 L = n * 0.08206 atm.L/mol.K * 298.15 K

n = 4.087*10^-2 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

use:

mass of CO2,

m = number of mol * molar mass

= 4.087*10^-2 mol * 44.01 g/mol

= 1.799 g

Answer: 1.80 g

d)

Given:

P = 1.0 atm

V = 1.0 L

T = 298.15 K

find number of moles using:

P * V = n*R*T

1 atm * 1 L = n * 0.08206 atm.L/mol.K * 298.15 K

n = 4.087*10^-2 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

use:

mass of H2O,

m = number of mol * molar mass

= 4.087*10^-2 mol * 18.02 g/mol

= 0.7364 g

Answer: 0.736 g