Answer:
2.83 g
Explanation:
At constant temperature and pressure, Using Avogadro's law
Given ,
V₁ = 2.12 L
V₂ = 3.12 L
n₁ = 0.120 moles
n₂ = ?
Using above equation as:
n₂ = 0.17660 moles
Molar mass of methane gas = 16.05 g/mol
So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g
<u>2.83 g are in the piston.</u>
Answer:
= 0.134;
= 0.866
The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr
Explanation:
For each of the solutions:
mole fraction of isopropanol () = 1 - mole fraction of propanol (
).
Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.
Furthermole, the partial pressure of isopropanol = *vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr
The partial pressure of propanol = *vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr
Similarly,
In the vapor phase,
The mole fraction of propanol () =
Where, is the partial pressure of propanol and
is the partial pressure of isopropanol.
Therefore,
= 5.26/(34.04+5.16) = 0.134
= 1 - 0.134 = 0.866