What is the mass of O2 in grams?

Question

In a mixture of Cl2 and O2 gases, the partial pressure of Cl2 is 0.150 atm. The total pressure is 0.365 atm. The temperature is 32.0oC. If the total volume of the gas mixture is 538 mL, what is the mass of O2 in grams?

Answer 1

Call p1 = p(Cl2), p2 = p(O2)

p1 = 0.15 atm, p = p1 + p2 = 0.365 atm

-> p2 = 0.365 - 0.15 = 0.215 atm

V = 0.538 l

T = 273 + 32 = 305 K

R = 0.082 L*atm / K

For a mixture, all gases share the same V and T:
p2.V = n2.RT -> n2 = p2.V/(R.T) = 0.004625 mol

Therefore, the mass of O2 is:
0.004625 * 32g = 0.148 g

Answer 2

the partial pressure of O2 is 0.365-0.150 = 0.185 atm
The number of moles of O2 is derived from the formula pV =nRT

where p is the pressure, V the volume, n number of moles, T absolute temperature T = t +273, R constant
so n = pV/RT R = 8.31 ;T= 273+32 = 305°K; p= 0.185 *10^5 Pa;
V= 0.000538 m^3
We find n = 0.0039 moles
As each mole of O2 weights 32g
we found m = 0.0039*32=

Answer 3

Is this college chem? I hate chem ever since i started college . . . sorry that didn't help you . . .

Answer 4

try using...

PV=nRT

in your case...
P=0.215atm
V=0.538L
n=number of moles O2
R= I think its 0.00825atm-L/mol-K... look it up
T= 305K