Answer:
Moles added = 0.0722 moles
Mass added = 2.3104 g
Explanation:
Given:
Pressure = 640.0 torr
The conversion of P(torr) to P(atm) is shown below:
[tex]P(torr)=\frac {1}{760}\times P(atm)[/tex]
So,
Pressure = 640 / 760 atm = 0.8421 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 42.9 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.8421 atm × 42.9 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 1.4759 moles
Some new moles have been added and the volume has increased to 45.0 L
Using Avogadro's law
[tex]\frac {V_1}{n_1}=\frac {V_2}{n_2}[/tex]
Given ,
V₁ = 42.9 L
V₂ = 45.0 L
n₁ = 1.4759 moles
n₂ = ?
Using above equation as:
[tex]\frac{42.9}{1.4759}=\frac{45.0}{n_2}[/tex]
[tex]n_2=\frac{45.0\cdot \:1.4759}{42.9}[/tex]
n₂ = 1.5481 moles
Moles added = n₂ - n₁ = 1.5481 moles - 1.4759 moles = 0.0722 moles
Molar mass of oxygen gas = 32 g/mol
So, Mass = Moles*Molar mass = 0.0722 * 32 g = 2.3104 g
Moles added = 0.0722 moles
Mass added = 2.3104 g
