In a container with a total of 1.25 mol of gas particles, the partial pressure o
ID: 1072632 • Letter: I
Question
In a container with a total of 1.25 mol of gas particles, the partial pressure of CH, is 80 torr and the partial pressure of O_2 is 420 torr. Find (i) the total pressure of the mixture, and (ii) the number of moles of each gas. In a sample containing 0.20 mol N_2 and 0 60 mol He, the partial pressure of N, is 30 atm. Find (i) the partial pressure of He and the (ii) total pressure of the sample. In an 80.0 kPa sample of NH, and H_2O the partial pressure of NH_3, is 16.0 KPa If there are 0.30moi NH, find the number of moles of in the sample.Explanation / Answer
b)
n = 25 mol
CH4 = 80 torr
O2 = 420 torr
a)
total P = P1+P2 = 80+420 = 500 torr
b)
mole of each gas
x-O2 = 420/500 = 0.84
mol = 0.84*25 = 21 mol of O2
for methane
mol = 25-21 = 4 mol of CH4
c)
m = 0.2 N2
m = 0.6 He
N2 = 3 atm...
find
Parital pressure of He
so...
He/N2 ratio is 0.6/09.2 = 3 times
so
P-He = 3*3 = 9 atm
ii
total P = 9+3 = 12 atm
d)
P = 16 kPa NH3
P H2O = ?
mol frac of NH3 = 16/80 = 0.2
so
0.2 fraction = 0.3 mol
0.3/0.2 = 1.5 total mol
then
mol of H2O = 1.5-0.3 = 1.2 mol of -H2O