At 1.00 atm, the solubility of O2 in water is 2.18 10-3 M at 0°C and 1.26 10-3 M at 25°C.?
What volume of O2(g), measured at 25°C and 1.00 atm, is expelled when 545 mL of water saturated with O2 is heated from 0 to 25°C?
I'm really lost, I tried this question two different ways and still seem to get it wrong. I used (moles= molarity x volume (.545) to find the moles at 0 C and 25 C and then found the difference between them and plugged in my numbers to the V=nRT/P and it was wrong. So any help would be great appreciated. THANX!
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I don't know of any way to do it other than what you said you've already tried. Try going through it again:
2.18e-3 (mol/L) x .545 L = 0.0012 moles at 0C
1.26e-3 (mol/L) x .545 L = 6.867e-4 moles at 25C
0.0012 - 6.867e-4 =5.133e-4
Use this as your moles in the ideal gas equation.
V = [(5.133e-4 mol) x (0.08206 L atm /mol K) x (298 K)] /1.00 atm
V= 0.01255 L or 12.55 mL
If you got this same answer and it was wrong, I don't know what else to tell you.
I hope this was helpful!
O2 In Water
I did by your way and got corect answer. thanks so much :)