Question

Calculate the molarity and molality of
49.0 wt% of HF.
How much solution is required to prepare
0.270 L of 0.302 M HF?

Answer 1

Can't remember how to do basic chemistry.

Answer 2

That's hot. Is it dissolved in water?

Answer 3

You can calculate the molality even if you don't know the solvent.
In 100 g solution,
49 g HF or 49/20 mol. HF
51 g solvent or 0.051 kg solvent.

Molality = Moles of solute/Mass of solvent in kg
= 49/20 * 1/0.051

Answer 4

Where's this 20 mol coming from?

Answer 5

49 g HF
1 mol HF ---> 20 g HF (1 g + 19 g)
49/20 mol HF ---> 49 g HF

Answer 6

I'm sorry, I still don't get the 20 g HF here.

Answer 7

That is to say that the molar mass of HF is 20 grams. We calculate molar masses of compounds using molar masses of the individual consisting atoms and adding them up. From my memory, H ---> 1 g/mol and F ---> 19 g/mol. Adding both, HF ---> 20 g/mol.

Answer 8

Sorry, I keep having to refresh so I can type here. I wrote it as 20 g/mol. Now how did you figure the solvent was 51 g?

Answer 9

OK, so if I just assume for a second that I have a 100 gram solution, it wouldn't affect the concentrations because I'm calculating the relative ratios.

Now I'm given a 49.0% by wt. HF solution. That means for every 100 g of solution, I have 49 g. The remaining 51 g is obviously the solvent.

Answer 10

Btw the answer in the txtbk is 2.87 mL.

Answer 11

What? Yes, that's the answer to "How much solution is required to prepare
0.270 L of 0.302 M HF?" but did it forget the first part, i.e., "calculate the molarity and molality of..."?

Answer 12

Let's get done with your confusion with molality before we proceed.

Answer 13

Ok.

Answer 14

So...

Answer 15

sorry, goddarn roommates decided to trash my room

Answer 16

Anyway I'll just show you why the answer is that. Assuming that the solution was made in water with density 1 g/ml, we can say that for every 100 g solution, it consists of 49 g HF and 51 g water.

Or since we know that 1 g water = 1 ml water, 51 g water is 51 ml of the solvent.

Now we can calculate the molarity = moles of solute per unit volume of solution (litres).

Again, as I said, 49 g HF is 49/20 mol. HF. And 51 ml is 0.051 L. Therefore, the molarity is \[M = \rm \frac{49/20 ~ mol}{0.051~L } = 48.04 M\]

Answer 17

We need to prepare a 0.27 L solution of 0.302 M HF. The thing in this type of statement is we have to work with the same HF we're given but we can add more solvent to reduce the concentration to 0.302 M instead of the original 48.04 M. So what do we do?

We need 0.302 * 0.27 = 0.0815 moles HF in this solution that we'd prepare, right? Now how much of the original solution contains 0.0815 moles given that 1 litre corresponds to 48.04 moles?

Answer 18

Sorry mate, I'm just really tired, and distracted by my roommates' previous actions. Its taking me a while to figure out molality.

Answer 19

OK, we'll start right where the problem begins. What's the solvent anyway? This problem depends a lot on what the solvent is.

Answer 20

It didn't say. I copied directly from the slide.

Answer 21

In that case, we assume water. If you're given 2.87 ml as the answer then a quick calculation suggests that 0.302*0.27/0.00287 = 28.41M is the molarity of the solution.

Answer 22

ISthe molarity and molality the same, but with different units?

Answer 23

They have different definitions.

Answer 24

Yes, I wrote that in my notes (Molarit- Number of moles of substance contained per liter of solution) while molality is the number of moles of substance contained per kilogram of solvent.

Answer 25

Values are the same though right?

Answer 26

they're not.

Answer 27

bugger

Answer 28

I just did a calculation, and backtracking from the answer they've given us, the density of the solvent they've given us is turning out to be 51/86 = 0.59 g/ml. That's not water.

Answer 29

Wait so how do I get 2.87 mL?