Question

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.

Answer 1

According to Dalton's law of partial pressures

Total pressure P_t = P_N2 + P_He

Where P_N2 = partial pressure of N2, P_He = partial pressure of He

There is a relation connecting partial pressure , total pressure and mole fraction.

P_N2 = x_N2 P_t

Where x_N2 is molefraction of N2.

P_He = x_He P_t

Where x_He is molefraction of He.

P_He = x_He P_t

x_He = 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

x_He = 4.6/(4.6+1.83) = 4.6/6.43 = 0.715

P_t = 40.1 atm

P_He = 0.715 * 40.1 atm = 28.67 atm

Patial pressure of He = 28.7