Three gases (8.00 g of methane, ch4, 18.0 g of ethane, c2h6, and an unknown amount of propane, c3h8) were added to the same 10.0-l container. at 23.0 ∘c, the total pressure in the container is 5.50 atm . calculate the partial pressure of each gas in the container.

The partial pressure for methane, Ethane and propane is 1.21 atm, 1.24 atm, 2.83 atm respectively.

Given here,

Mass of methane 8g

Mass of ethane is 18g

From ideal gas formula

PV = nRT

(5.50 atm)(10 L) = n(0.0821 )(296 K)

n = 2.263 mol

Moles methane = 0.499 mol

Moles ethane = 0.599 mol

Moles propane = 1.165 mol

According to Raoult's law

where,

Pa is partial pressure

P' is total pressure

Xa is mole fraction

By solving the equation for each

We can conclude that the partial pressure for methane, Ethane and propane is 1.21 atm, 1.24 atm, 2.83 atm respectively.

To know more about Raoult's Law, refer to the link:

link

idk what the question is about