Question

What mass (in g) of oxygen must be added to 8.89 g of neon at 23.6oC, and in a 103 L contain for a final pressure of 850 mmHg?

Answer 1

850 mm Hg = (850/760) atm = 1.12 atm, T = 296.75 K , V = 103 L

using PV=(n1 + n2) RT

n1 + n2 = PV/RT

(8.89/ 20) + n2 = (1.12 X 103 /0.0821 X 296.75)

n2 = 4.735 - 0.445 = 4.29 mol

hence amount of oxygen = 4.29 X 32 = 137.3 g

you cana also do it by other way by using

PT = x_Ne P⁰ Ne + x_O2 P⁰_O2

where x = mole fraction of the gas , and P⁰ =vapour pressure of the pure gas

from this we can find the mole fraction , hence moles of oxygen needed. provided we are knowing the P⁰ of the gases.