Calculating Volume of 0.835M Na2SO4 Solution (0.0534 mol Na+)

[RJLiberator]

Question: What volume of a 0.835 M solution of Na2SO4 do you have if there are 0.0534 mol of Na+ present in the solution?

What I did: Since we have 0.0534 moles of Na+ ions in the Na2SO4 molecule, I used that for the equation. Since 0.0534 moles is in a 1:1 ratio with Na2SO4 I can conclude we have 0.0534 moles of Na2SO4.

M = Moles/Volume
0.835M = 0.0534moles of Na2SO4/V
Solving, I get 6.40x10^-2L

Is my reasoning correct?

 

[Simon Bridge]

Well that is the definition of molarity you used there... so you reasoned:

M=n/V => V=n/M plug numbers in.
No worries.

 

[Borek]

Since 0.0534 moles is in a 1:1 ratio with Na2SO4 I can conclude we have 0.0534 moles of Na2SO4.

Can you? Is it really 1:1?

 

[RJLiberator]

Ah, prior to my test I spoke with my professor and it appears that I can do no such thing. There is 2 Na ions and 1 So4 ions so it is actually a 2 to 1 ratio. I will work this out when I get home tonight. :)

Thank you.

 

[DeeNos]

I would say that your reasoning and calculations are correct. In order to calculate the volume of a solution, we use the equation M=moles/volume. In this case, we know the molarity (0.835M) and the number of moles of Na+ ions (0.0534 mol), so we can rearrange the equation to solve for volume. By substituting the values and solving, we get a volume of 6.40x10^-2 liters. This means that in 1 liter of the 0.835M Na2SO4 solution, there are 0.0534 moles of Na+ ions. Great job on your calculations!