Question

If the sollubility of sodiym acetate (Molar Mass = 82 g/mol) is 76 grams per 100 grams of water, which of the following solutions would be considered supersaturated (holding more than the normal maxiym solution)? a.) 8.5 moles of sodium acetate dissolved in 1 L of water, b.) 1.8 moles of sodiym acetate dissoved in 300 mL of water, c.) 5.5 moles of sodiym acetate dissolved in 500 mL of water, or d.) 1.2 moles of sodium acetate dissoved in 200 mL of water? I would like to understand how to calculate these kinds of problems. Please explain step by step how I should go about solving this so that I may learn how to do it for myself. There is an answer on Chegg for this already but it doesn't appear to be accurate nor does it clarify what is going on for "understanding" sake as I tried to solve it for myself using the method there and it was more confusing than helpful. So, I am asking it an effort to gain a deeper and more accurate understanding.

Answer 1

First, calculate the solubility as mole per gram (or mL) of water, and mole per L of water. This will help you know if the solution are supersaturaded or not.

To calculate the moles we have: moles = ,/MW
moles = 76 g/82 g/mol = 0.93 moles/100 g of water.

Now let's change the mass of water to volume, the density of water, will be assumed as 1 g/mL. so 100 g of water is 100 mL according to this.

Then the solubility of sodium acetate in mL of water and L of water will be (assuming 100 mL is 0.1 L)
S = 0.93/100 = 0.0093 moles/mL of water
S' = 0.93/0.1 = 9.3 moles/L of water

Now we have the solubility of sodium acetate in mole per mL of water and in mol per L of water. Now let's see all cases:

a) 8.5 moles/L

In this case if we compare the S' to this solubility, we see that 8.5 mol/L is smaller than 9.3 moles/L, so this means that this solution is soluble but insaturated (less solute that it can hold).

b) 1.8 moles/300 mL
Let's see first the rule of 3:

0.93 moles ---------> 100 mL
X moles ------------> 300 mL

X = 300 *0.93 / 100 = 2.79 moles per 300 mL should be dissolved to be a saturated solution, and once again, we have 1.8 moles, that's less of the minimum quantity required to have a saturated solution. So this is another insaturated solution.

c) 5.5 moles per 500 mL
We'll do the same procedure as before:
0.93 ----> 100 mL
X ------> 500 mL
X = 500 * 0.93 / 100 = 4.65 moles per 500 mL

In this case we can see that according to the real solubility of sodium acetate, in 500 mL we can only have a maximum of 4.65 moles, and this solution have 5.5 moles, we are in excess of solute, so this solution is supersaturated.

d) 1.2 moles per 200 mL
0.93 moles -----> 100 mL
X -------> 200 mL
X = 200 * 0.93 / 100 = 1.86 moles

Once again, we have less of the minimum quantity (that is 1.86 moles in 200 mL), so this solution is insaturated.

Only solution c) is supersaturated.

I hope you understand better now. Hope this helps you