A 51.3-g sample of N2 (g) is mixed with 18.4 g of He (g) in a container. If the total pressure is 40.1 atm, what is the partial pressure of He? Only enter your numerical answer. Do not enter units.
According to Dalton's law of partial pressures
Total pressure Pt = PN2 + PHe
Where PN2 = partial pressure of N2, PHe = partial pressure of He
There is a relation connecting partial pressure , total pressure and mole fraction.
PN2 = xN2 Pt
Where xN2 is molefraction of N2.
PHe = xHe Pt
Where xHe is molefraction of He.
PHe = xHe Pt
xHe = no.of moles of He/(total no.of moles)
no.of moles of He = mass in grams/molar mass = 18.4 g/4 g/mol = 4.6 mol
no.of moles of N2 = mass in grams/molar mass = 51.3g/28 g/mol = 1.83 mol
xHe = 4.6/(4.6+1.83) = 4.6/6.43 = 0.715
Pt = 40.1 atm
PHe = 0.715 * 40.1 atm = 28.67 atm
Patial pressure of He = 28.7
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