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Chemistry Forums for Students => High School Chemistry Forum => Chemistry Olympiad and other competitions => Topic started by: yuheng_wu on December 16, 2020, 02:45:48 PM

Given:
 dioxide is being disolved in water.
 c=8mg/l
 V_{water}=10,0 liters
 T = 25 °C = 298,15K
 p = 1013 hPa
The question is: 'how much volume dioxide kan disolve maximally in 10,0 liters of water at a temperature of 25°C and pressure of 1013hPa?
The answer is 61,1mL, however when I solve this problem my answer is 6,11mL. Could someone please tell me what I'm doing wrong?
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First, O_{2} is dioxygen, not "dioxide".
You are using the concentration in mg/L, and working out the volume that dissolves in 1L. You are asked for the volume that can dissolve in 10L.

First, O_{2} is dioxygen, not "dioxide".
You are using the concentration in mg/L, and working out the volume that dissolves in 1L. You are asked for the volume that can dissolve in 10L.
So c= n_{dissolved substance}/V_{substance} here I can say that V _{substance} = 10 liters? I though that this is incorrect since V _{substace} is equal to V_{water}+V_{dioxygen} = 10 liters + x liters or am I understanding this formula incorrectly? I do understand my mistake now but I don't understand this formula.

Such a small amount of oxygen dissolved in water practically does not increase its volume.
You have calculated the volume of oxygen dissolved in 1 L of water in the longest possible way. How much oxygen will there be in 10 liters of water?

Such a small amount of oxygen dissolved in water practically does not increase its volume.
You have calculated the volume of oxygen dissolved in 1 L of water in the longest possible way. How much oxygen will there be in 10 liters of water?
Ah yes the volume of oxygen would not really increase it's volume indeed, I understand. You said I have calculated the volume in a very inefficient way? How could I do this faster/more efficiently please?

10·(0.008/32)·22400·(298.15/273.15)=

10·(0.008/32)·22400·(298.15/273.15)=
Thank you very much.