Given that the solubility of sodium chloride is 36 grams per 100 grams of water. Which of the following solutions would be considered supersaturated?

A. dissolve 5.8 moles of NaCl in 1 L of water

B. dissolve 1.85 moles of NaCl in 300 ml of water

C. dissolve 3.25 moles of NaCl in 500 ml of water

D. none of the above

My husband and I are arguing...I think that the answer is A and he thinks it's C--please help and explain. Thanks.

It's a tie between you and your husband. I don't think either is correct.

I looked up the molar mass of NaCl and found it to be 58.442.
So A is 5.8 x 58.442 = about 338 g
B is 1.85 x 58.442 = 108.11 g
C is 3.25 x 58.442 = 189.99 g
D--

How much can dissolve.
A is 36 x 1000/100 = 360 g so 338 should dissolve easily.
B is 36 x 300/100 = 108 which is close and I might call it saturated but not supersaturated.
C is 36 x 500/100 = 180 g and it can dissolve nearly 190 g which eliminates C. I pick D. Check my thinking.
One note: Note that the problem states "per 100 GRAMS water" while the answers are in L and mL. I have assumed the density of water is 1.00 g/mL which makes 1 mL have a mass of 1.00 gram. I don't know if this was an attempt to throw you off or not. Also, some profs may argue that we shouldn't use that many places in the molar mass of NaCl. However, if we use 58.4 (three significant figures to agree with 3 s.f. for the moles) then 1.85 x 58.4 = 108 which still makes a saturated solution. While we are playing these IF games, we might argue that the 300, 500, and 1 L contain only 1 s.f., THEN, some of the conclusions might change. However, I think the intent of the problem is for the answer to be none of these.

Dr Bob,

First let me thank you for your help and explanations, so that I can understand what I'm supposed to be doing--it's much appreciated!

I have a question though, at what point do I know that a solution would be saturated, supersaturated, unsaturated, etc. I know the word definitions for these, but how do you know that if your answer for option "b" above was 108.11, then how do you know that it is not saturated, or supersaturated? I know that supersaturated would mean that the substance would not completely dissolve, but how do you tell in regards to numbers? I'm so confused! :-(

First, let me apologize because I goofed and not just a little but a lot. The correct answer is not D.

Unsaturated means that the solute has dissolved and more can dissolve.
Saturated means that the solute has dissolved and that one more crystal will not dissolve.
Supersaturated means (and this may sound confusing) that more of the solute has dissolved than normally possible. How can that happen? We can dissolve the solute in a hot solvent, then very carefully cool the solvent to room temperature. SOME solutions will hold all of the solute without any of it precipitating. The usual process is for it to precipitate (crystallize) but SOME solutes will stay in solution when, technically, they shouldn't. (In the case of a supersaturated solution, one can scratch the sides of the test tube with a stirring rod, or shake the solution, or drop in a crystal to act as a seed and the excess solute crystallizes almost instant. I made the error because I was trying to go back and forth between the question and my typing and I reversed some of the numbers.
Here are the answers (with my comments in bold) again
So A is 5.8 x 58.442 = about 338 g
B is 1.85 x 58.442 = 108.11 g
C is 3.25 x 58.442 = 189.99 g
D--

How much can dissolve.
A is 36 x 1000/100 = 360 g so 338 should dissolve easily.This is OK.

B is 36 x 300/100 = 108 which is close and I might call it saturated but not supersaturated. So we have dissolved 108.11 and it can hold 108 so this could be supersaturated depending upon how you want to handle the significant figure thing. I think the spirit of the question calls for you to decide this is a saturated solution.

C is 36 x 500/100 = 180 g and it can dissolve nearly 190 g which eliminates C. I pick D. This is the correct answer. The solvent CAN hold 180, but the 3.25 M tells us it HAS almost 190 g dissolved in it (note where I reversed the numbers in my mind) so it is clearly supersaturated. It has more solute (190 g) than the solvent can hold normally (180 g).
I hope this answer is sufficient and I certainly hope I didn't confuse you with my original answer. You were wise to ask me to explain it again--I noted the error when I started typing. Thanks for the heads up.

To determine which of the given solutions is supersaturated, we need to compare the amount of sodium chloride dissolved in each solution with the solubility of sodium chloride in water.

The given solubility of sodium chloride is 36 grams per 100 grams of water. We can use this information to convert the solubility to moles per liter, which is more convenient for comparing with the given solutions.

1. Convert the grams of sodium chloride to moles:
36 grams NaCl * (1 mole NaCl / 58.44 grams NaCl) = 0.617 moles NaCl

2. Convert the amount of solute to moles per liter:
(0.617 moles NaCl / 100 grams water) * (1000 grams water / 1 L water) = 6.17 moles NaCl/L

Now let's evaluate each solution:

A. Solution: dissolve 5.8 moles of NaCl in 1 L of water
The amount of solute (5.8 moles NaCl) is greater than the calculated solubility (6.17 moles NaCl/L). Therefore, this solution is supersaturated.

B. Solution: dissolve 1.85 moles of NaCl in 300 ml of water
To compare this solution with the solubility, we need to convert the volume to liters:
300 ml * (1 L/1000 ml) = 0.3 L
The amount of solute (1.85 moles NaCl) is less than the calculated solubility (6.17 moles NaCl/L). Therefore, this solution is not supersaturated.

C. Solution: dissolve 3.25 moles of NaCl in 500 ml of water
To compare this solution with the solubility, we need to convert the volume to liters:
500 ml * (1 L/1000 ml) = 0.5 L
The amount of solute (3.25 moles NaCl) is less than the calculated solubility (6.17 moles NaCl/L). Therefore, this solution is not supersaturated.

Based on the comparison, the correct answer is D. None of the given solutions are supersaturated.