In a mixture of O2 and Ne gases, the partial pressure of O2 is 0.135 atm.?

Question

The total pressure is 0.405 atm. The temperature is 26.0 oC. If the total volume of the gas mixture is 406 mL, what is the mass of Ne in grams?

Answer 1

double than O2

Answer 2

The partial pressure of Ne is:

p = 0.405 - 0.135 = 0.27 atm

From the state equation you can find the moles of Ne:

p*V = n*R*T, n = p*V/R*T, n = 0.27*0.406/0.082*(26 + 273), n = 4.47x10^(-3) mol

The atomic mass (or atomic weight) of Ne equals Ar = 20, so:

m = n*Ar, m = 4.47x10^(-3)*20 = 0.089 g (approx)