Question

A)A mixture of xenon and helium gases, at a total pressure of 851 mm Hg, contains 30.4 grams of xenon and 0.519 grams of helium. What is the partial pressure of each gas in the mixture?

P_Xe =  mm Hg
P_He =  mm Hg

B)A mixture of helium and argon gases contains helium at a partial pressure of 487 mm Hg and argon at a partial pressure of 471 mm Hg. What is the mole fraction of each gas in the mixture?

X_He =
X_Ar =

Answer 1

According to Dalton's Law of partial pressure , Partial pressure of gas in mixture = X _gas ( P total )

Where X _gas is mole fraction of gas.

First calculate no. of moles & mole fraction of each gas.

We know that , No. of moles = Mass / Molar mass

$\therefore$ No. of moles of Xe ( n _Xe ) = 30.4 g / ( 131.30 g /mol ) = 0.2315 mol

$\therefore$ No. of moles of He ( n _He ) = 0.519 g / ( 4.00 g / mol ) = 0.1298 mol

Mole fraction of Xe = Moles of Xe / Total moles of gas = n _Xe / ( n _Xe + n _He )

$\therefore$ Mole fraction of Xe = 0.2315 / ( 0.2315 + 0.1298 ) = 0.641

Mole Fraction of He = n _He / ( n _Xe + n _He )

$\therefore$ Mole fraction of He = 0.1298 / ( 0.2315 + 0.1298 ) = 0.359

Partial pressure of Xe = 0.641 ( 851 mm Hg ) = 545 mm Hg

Partial pressure of He = 0.359 ( 851 mm Hg ) =306 mm Hg

ANSWER : P Xe = 545 mm Hg & P He = 306 mm Hg

PART 2

First calculate total pressure of gas

P total = P He + P A r

Therefore, P total = 487 mm Hg + 471 mm Hg = 958 mm Hg

According to Dalton's Law of partial pressure , Partial pressure of gas in mixture = X _gas ( P total )

Where X _gas is mole fraction of gas.

Therefore, X _gas = Partial pressure of gas / P total

$\therefore$ X _He = 487 mm Hg / 958 mm Hg = 0.508

X _Ar = 471 mm Hg / 958 mm Hg = 0.492

ANSWER : X _He = 0.508 & X _Ar = 0.492