Answer:
Explanation:
Given data:
A) Volume of CO₂ = 0.100 L
Pressure = 307 torr
Temperature = 26°C
Mass of CO₂ = ?
Solution:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
26+273 = 299 K
307 torr/760 = 0.40 atm
0.40 atm × 0.100 L = n × 0.0821 atm.L/ mol.K × 299 K
0.04 atm.L = n × 24.55 atm.L/ mol
n = 0.04 atm.L / 24.55 atm.L/ mol
n = 0.002 mol
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.002 mol × 44 g/mol
Mass = 0.088 g
Given data:
b) Volume of C₂H₄ = 8.75 L
Pressure = 378.3 Kpa
Temperature = 483 K
Mass of C₂H₄ = ?
Solution:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
378.3KPa/101 = 3.73 atm
by putting values
3.73 atm × 8.75 L = n × 0.0821 atm.L/ mol.K × 483 K
32.64 atm.L = n × 39.65 atm.L/ mol
n = 32.64 atm.L / 24.55 atm.L/ mol
n = 0.82 mol
Mass of C₂H₄ :
Mass = number of moles × molar mass
Mass = 0.82 mol × 28 g/mol
Mass = 22.96 g
c)
Given data:
A) Volume of Ar = 221 mL
Pressure = 0.23 torr
Temperature = -54°C
Mass of Ar = ?
Solution:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
-54+273 = 219 K
0.23 torr/760 = 0.0003 atm
221 mL/1000 = 0.221 L
0.0003 atm × 0.221 L = n × 0.0821 atm.L/ mol.K × 219 K
6.63×10⁻⁵ atm.L = n × 17.98 atm.L/ mol
n = 6.63×10⁻⁵ atm.L / 17.98 atm.L/ mol
n = 0.37×10⁻⁵ mol
Mass of Ar:
Mass = number of moles × molar mass
Mass = 0.37×10⁻⁵ mol × 39.9 g/mol
Mass = 14.76 ×10⁻⁵g
