Calculating Water Volume from Molar Concentration of NaCH3COO

[Nicole1998]

Homework Statement

2.8 g of NaCH₃COO is dissolved into an unknown volume. The resulting concentration produced is 4.3 * 10^-2 M, what volume of water was used?

Homework Equations

c = n/V

The Attempt at a Solution

I[/B] calculated molar mass of the NaCH₃COO as 82.0 g/mol, which is 0.034 mol. I then used the c = n/V equation.

4.3 * 10^-2 M = 0.034 mol / V
V = 0.79 L

Where I am confused now is if that is the volume of the solution, or of the water? I think it's the volume of solution, and am not sure how to calculate the water.

 

[Quantum Defect]

Homework Statement

2.8 g of NaCH₃COO is dissolved into an unknown volume. The resulting concentration produced is 4.3 * 10^-2 M, what volume of water was used?

Homework Equations

c = n/V

The Attempt at a Solution

I[/B] calculated molar mass of the NaCH₃COO as 82.0 g/mol, which is 0.034 mol. I then used the c = n/V equation.

4.3 * 10^-2 M = 0.034 mol / V
V = 0.79 L

Where I am confused now is if that is the volume of the solution, or of the water? I think it's the volume of solution, and am not sure how to calculate the water.

You are correct that molarity refers to moles solute per volume of solution. You could estimate a volume of water in a number of ways:
Estimate that the volume contribution from the sodium acetate is small compared to the volume of the water and say that the volume of water is the same as the volume of solution. You could estimate the total volume if you estimated that the total volume was the combined volume of sodium acetate plus volume of water (you could calculate the volume of sodium acetate if you looked up the density of the sodium acetate in the CRC Handbook. If you wanted to calculate the volume of water exactly, you would need to know the partial molar volumes of each species at this concentration.

Now, think about how much precision you have in your numbers, and think about which method of approximation is appropriate. [The percent error in the weight of sodium acetate is 3% and the percent error in the molarity is 2% -- the percent error in the volume will be 3%]. V = 790 mL +/- 30 mL

Since the uncertainty in the volume is pretty big, what kind of assumptions would you be safe making in order to calculate the volume of water used?

 

[Chestermiller]

If you knew the density of a 0.043 molar solution of NaCH3COO in water, then you would know the total number of grams of solution in 0.79 L. Since there are 2.8 gm of NaCH3COO, the rest would be water. You could then look up the density of water at that temperature, and determine the volume of the water before the solution was mixed.

Chet

 

[Borek]

At 20°C density of the 0.043 M sodium acetate solution is 1.0000 g/mL (that's a little bit higher than pure water, 0.9982 g/mL at the same temperature).

In general, you can safely assume density of most solutions to be 1 g/mL. That's not exactly true, but it is better than nothing when you have no other data. And before you will be forced to use density tables, you will probably get some intuition about when it matters.

 

[DeeNos]

Your calculation is correct. The volume calculated is the volume of the final solution, which includes both the NaCH3COO and the water used to dissolve it. In order to determine the volume of water used, you would need to know the initial volume of water before the NaCH3COO was added. If the initial volume of water is unknown, it is not possible to calculate the volume of water used.