Question

A mixture of carbon dioxide and neon gases, in a 8.59 L flask at 56 °C, contains 12.8 grams of carbon dioxide and 1.86 grams of neon. The partial pressure of neon in the flask is_____ atm and the total pressure in the flask is ______atm.

Answer 1

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

Molar mass of Ne = 20.18 g/mol

n(CO2) = mass of CO2/molar mass of CO2

= 12.8/44.01

= 0.2908

n(Ne) = mass of Ne/molar mass of Ne

= 1.86/20.18

= 0.0922

n(CO2),n1 = 0.2908 mol

n(Ne),n2 = 0.0922 mol

Total number of mol = n1+n2

= 0.2908 + 0.0922

= 0.383 mol

we have:

V = 8.59 L

n = 0.383 mol

T = 56.0 oC

= (56.0+273) K

= 329 K

we have below equation to be used:

P * V = n*R*T

P * 8.59 L = 0.383 mol* 0.08206 atm.L/mol.K * 329 K

P = 1.2037 atm

p(Ne),p2 = (n2*Ptotal)/total mol

= (0.0922 * 1.2037)/0.383

= 0.2897 atm

partial pressure of Neon = 0.290 atm

total pressure = 1.20 atm