Answer:
The concentration of O₂ in the sample is 5,1x10⁻³ M (mol/L)
Explanation:
To obtain the moles of gas you should use ideal gas formula:
PV/RT = n
Where:
P is pressure: 650 mmHg or 650 mmHg* ( 1 atm / 760 mmHg) = 0,855 atm
V is volume: 10 L
R is gas constant: 0,082 atm·L/mol·K
T is temperature: 160°C or 160 + 273,15 = 433,15 K
And n is mol number
Thus, replacing bold values in ideal gas formula:
n = 0,24 mol
The mole fraction of O₂ in atmospheric air is 0,21. Thus, the moles of O₂ there are:
0,24 moles of atmospheric air ₓ 0,21 = 0,051 O₂ moles
There are many ways to express concentration but we will express it in molarity (mol of solute / L of solution) understanding solute as O₂ and solution as atmospheric air. So:
0,051 O₂ moles / 10 L = 5,1x10⁻³ M -<em>M is molarity</em>-
I hope it helps!
