Question

A mixture containing 0.767 mol He(g), 0.331 mol Ne(g), and 0.113 mol Ar(g) is confined in a 10.00-L vessel at 25 ∘C.

1.Calculate the partial pressure of He in the mixture.

2.Calculate the partial pressure of Ne in the mixture.

3.Calculate the partial pressure of Ar in the mixture.

4.Calculate the total pressure of the mixture.
*all answers should be in ATM

Answer 1

Dalton's law: According to Dalton's law the total pressure of gases mixture (gases of mixture do not react chemically) is equal to the sum of all the partial pressure of the component gases of mixture at constant temperature and volume Given that 0.767 mol He(g) 0.331 mol Ne(g) 0.113 mol Ar(g) Volume 10.00 L Here, is the total pressure of mixture of gases, p is the partial pressure of gas 1, , is the partial pressure of gas 2 and P, is the partial pressure of gas 3 Tem 25 OR 298 K Total moles of gas = 0.767 + 0.331 + 0.113 = 1.211 moles Mole fraction = number of mole s/ total moles Partial pressure : The pressure exerted by an individual gas in a mixture is known as its partial pressure. mole fraction He = 0.767/1.211 0.633 mole fraction Ne = 0.331/1.211 = 0.273 Mole fraction: mole fraction Ar = 0.113/1.211 = 0.0933 Mole fraction is a ratio of mole of solute or solvent to the total number of moles (mole of solute + moles of solvent) or number of mole of solution Total pressure P = nRT/V = (1.211) (0.082) (298)/10 2.96 atm Raoult's law: PP He 0.633 x 2.96 atm = 1.87 atm PP Ne 0.273 x 2.96 atm=0.808 atm According to the Raoult's law in the mixture the partial pressure of a gas is equal to the product of its mole fraction and the total pressure of PP Ar 0.0933 x 2.96 atm = 0.276 atm gases: Here; P partial pressure of gas A, : Vapor pressure of mixture and x is the mole fraction of gas A