[mathieumg]

A solid is composed of NaNO₃ and Mg(NO₃)₂. When 6.50 g of this solid is dissolved in 50.0g of water, the freezing point drops by 5.40 ^oC. What is the composition (in mass), of this solid?

I said that:

ΔT = 5.40 ^oC = 1.86^oC * Kg/mol * m * i

I assumed that i = 1 but I am not sure about this and I believe in may be one of the sources of my mistake.

I isolated m to get:

m = 2.903 mol/Kg

Then found the amount of mols:

2.903 mol/Kg = n_solute/0.05 Kg H₂O

n_solute = 0.145 mol

Then I established the following equation:

x NaNO₃ * mol NaNO₃/85.00 g NaNO₃ + (6.50g-x) Mg(NO₃)₂ * mol Mg(NO₃)₂/148.33 g Mg(NO₃)₂ = 0.145 mol

Then I isolate x (where x is the amount of NaNO₃ in grams):

x = 20

Which would mean I have 20g of NaNO₃ and a negative amount of Mg(NO₃)₂, and that doesn't make any sense at all.

I am seeking help with this problem, I don't see what I'm doing wrong.

The answers given are 4.11g of NaNO₃ and 2.40g of Mg(NO₃)₂.

Thank you very much in advance

[Borek]

i is not 1, it has different values for both salts.

Think how to get two equations in two unknowns from your data.

[mathieumg]

I don't know how to handle that equation with multiple i's.

Is it something like ΔT = 5.40 oC = 1.86oC * Kg/mol * m * (i₁ + i₂) or I'm in the field?

I need some guidance here.

Thank you very much

[Borek]

m₁*i₁+m₂*i₂

[Ayesh]

Van't Hoff Factor (i) = # particules/ # molecules

Examples:

NaCl -> Na+ + Cl-
             2  for  1      then i=2

Na2So4 -> 2Na+ + SO4-2
                  3    for   1         then i=3

[mathieumg]

So the i for NaNO₃ would be 4?

The i for Mg(NO₃)₂ would be 9?

[Borek]

So the i for NaNO₃ would be 4?

Why? How does this salt differ from NaCl? It dissociates into two separate ions.

Are you squaring them, or what?

[mathieumg]

Chemical equations have always been my major weakness, I would need some enlightmen as to how to proceed exactly.

[pfnm]

How I'd proceed:

Work out what you have (what the question has given you), work out what the question is asking for (what answer you need to give).

This is where equations come in. Often you'll have a number of ways to answer a question.

Here I can see two methods. Let's try both.

First, what has the question told us?

We have 6.50g of a solid containing only Mg(NO3)2 and NaNO3. The molar mass of our sample, then, is the sum of the molar masses of Mg(NO3)2 (148.31g/mol) and NaNO3 (84.99g/mol). This is 233.3 g/mol.

We now have a molar mass and a sample mass. Divide sample mass by molar mass to get the number of moles of our sample solute.

6.50g sample/233.3g/mole molar mass sample = 0.0279 moles of the solute.

Now, we can go two ways here:

Look at Mg(NO3)2, and NaNO3. If you know your ions, you'll know Mg(NO3)2 will dissociate into 3 ions: Mg2+, and 2NO3-. Similarly NaNO3 will dissociate into 2 ions, Na+ and NO3-. That gives us a total of 5 ions.

Let's look at our freezing point depression equation.

ΔT = Kf x m x i

We have ΔT, Kf, and now we have i.

i, the van't Hoff factor, for most ionic compounds = the number of discrete ions in a formula unit of the substance. For example, NaCl, 2 ions. NaNO3, 2 ions, Ca3(PO4)2 (calcium phosphate), 5 ions, Cu(ClO3)2 (copper (II) chlorate), dissociates into3 ions, etc.

The van't Hoff factor for multiple electrolytic ionic substances, is the sum of the individual ionic ions present in an aqueous solution.

So we work on our equation:

(with working)

5.40oC = 1.86oC/m x m x 5

5.40oC = 9oC/m x m

molality = 5.40oC / 9oC/m

Simply divide 9 under 5.40, to isolate the unknown. Cancel the oC degrees. You'll be left with 0.6m.

Molality is equal to moles of solute component divided by kilograms of solvent.

0.0279 moles/0.050kg H2O = 0.558m, close to 0.6.

0.6x0.05 gives us moles: 0.03 moles, close enough to our original calculation of 0.0279. We will of course stick with the lower number.

Next, since we know we have 5 ions present, and we know Na+, and Mg2+ will account for 2 of those ions, leaving NO3- to account for the rest (2 of the ions come from Mg(NO3)2 and 1 of the ions comes from NaNO3).

Looking back at the question, they want the mass of the two components. To find this, we must think for a second. How many moles of ions are in solution?

The TOTAL answer, of course, is 5 x 0.0279 moles. Since the sample was 0.0279 moles, each single constituent ion going into solution will also have the same weight. We don't need the total answer yet, though. We need specific answers: how much NO3-, how much Mg2+?

Easily found: since there are 0.0279 moles of Mg2+ ions in solution, we multiply moles by molar mass (24g/mol Mg2+). The answer is 0.669 grams.

Two NO3- ions complete the ionic formulation of Mg(NO3)2. So we times 0.0279 moles by two, equalling 0.0558 moles and 0.0279x62g/mol NO3- ions for 3.46g NO3- ions.

Add the weight in grams of the magnesium with it's corresponding nitrate. You ought to get 4.13g or thereabouts.

6.50g - 4.13g = 2.37 g, the weight of the other solute, NaNO3.

----

Hopefully you can see what I've done there, and if you cannot, I'll be happy to explain further - but before I sign off, there's a much quicker way for this calculation. Be warned, though, if you don't understand freezing point depression equations, it would be good to 'solidify' the application of the equation in your memory by doing a few calculations like this, but the 'long way' if two ways are visible.

The short cut I'm talking about is this;

We know the moles of solute and the molar mass (add molar mass of NaNO3 and Mg(NO3)2). We know that NaNO3 will dissociate into Na+ ions and NO3 ions at a 1:1 ratio and that Mg(NO3)2 will dissociate into ions at a 1:2 ratio (Mg:NO3). So a total of five ions in solution exist (three NO3 ions, an Na+ ion and an Mg+ ion).

Obviously, the sample dissociates into ions of 0.0279 moles for each (3x0.0279 for NO3).

So, just by using molar mass of ions, moles of ions, we can work out the weight, in grams, of the constituent compounds of the sample, the same way as we just did earlier.

This question may be intended to mislead but if you play around and find ways of approaching it, you will be fine. Just sit and think about it, hard, if you can't see any approaches - but since I've just gone over everything, the method of solving the question should be clear.

---

Lastly:

You can also double check our calculations by solving the equation in another way:

ΔT = Kf x m x i

5.40oC = 1.86oC x m x i

We have 0.0279 moles of solute. We have 0.050kg of solvent.

molality = moles solute divided by kg solvent. This equals 0.558m.

1.86x0.558=1.038

5.40/1.038 = 5.2

so i = 5.2

This accords with our first result (0.6m and 5i, compared to 0.558m and 5.2i).

Theoretically, ionic substances like NaNO3 dissociate completely, giving clean van't Hoff factors for calculation. Experimentally, however, we might get a van't Hoff factor that is not a whole number.

So 5.2 is an accurate result, and if we solve the equation using this van't Hoff factor we will get the same results.

--

I know this is long, and I know I've given you all the answers and we're not supposed to do that, but I figure you had the answers anyway, you just needed to know the process. There are a few ways of approaching that sort of problem, but hopefully what I've written helps,

if not let me know,

Pete

[sjb]

How I'd proceed:

...

Here I can see two methods. Let's try both.

First, what has the question told us?

We have 6.50g of a solid containing only Mg(NO₃)₂ and NaNO₃. The molar mass of our sample, then, is the sum of the molar masses of Mg(NO₃)₂ (148.31g/mol) and NaNO₃ (84.99g/mol). This is 233.3 g/mol.
...

I'm not sure this is a valid jump to make. What is the definition of a mole (in this case?)

[Borek]

I'm not sure this is a valid jump

And I am sure it is not

[mathieumg]

Thank you for the help guys!