Molarity problems (help`~~~)

[joonkimdds]

How many grams of BiCl3 are needed to make 500.0 g of a 0.10-m aqueous solution?
I don't know how to do this.
The answer is 15.28g



What is the molarity of a 3.21 m KOH(aq) solution? (density of solution = 1.163 g/mL)
what I tried was
M = 3.21 mol / ? L of solution
we got 1163g solution which is eqal to 1163ml = 1.163L
but 3.21/1.163 = 2.76 and that's not the right answer.
The correct answer is 3.16M.

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[ZenoVT]

1)

m = molality = moles of solute/ kg of solvent

so .10 = x / .500
.050 mol = x

.050 mol x (209+ 3x35.5) = 15.75g

 

[joonkimdds]

1)

m = molality = moles of solute/ kg of solvent

so .10 = x / .500
.050 mol = x

.050 mol x (209+ 3x35.5) = 15.75g

at first, that's what I thought, but the very last step they used was
15.767 = 1.0315x
x = 15.28

 

[ZenoVT]

2)
this one is as simple as multiplying the two numbers

3.21 m x 1.163 g/mL = 3.73 M

 

[ZenoVT]

where are u getting hta # from? I'm getting a bit off for question #2 too

 

[joonkimdds]

The solution for #2 says
g of solution = g of solvent + g solute = 1000g + 180.1g = 1180.1g
and then they divide it by 1163 to get 1.015L.
M= mol KOH/L solution = 3.21mol KOH / 1.015L = 3.16M

even after reading the solution, I am still confused.

 

[tranv117]

The solution for #2 says
g of solution = g of solvent + g solute = 1000g + 180.1g = 1180.1g
and then they divide it by 1163 to get 1.015L.
M= mol KOH/L solution = 3.21mol KOH / 1.015L = 3.16M

even after reading the solution, I am still confused.

molality is equal to mol solute/kg solvent and Molarity is mol solute/L solution. so they calculated total grams of solution that you have and divide it by the density to get total liter of solution.

The key here is that Molarity looks at the volume of solution meaning the volume of solute + volume of the solvent while Molality is the mass of the solvent.

 

[joonkimdds]

For #1, this is how they started.
0.1m = (x/315.34)/ [(500-x)/1000]
confusing... 🙁

molality is equal to mol solute/kg solvent and Molarity is mol solute/L solution. so they calculated total grams of solution that you have and divide it by the density to get total liter of solution.

The key here is that Molarity looks at the volume of solution meaning the volume of solute + volume of the solvent while Molality is the mass of the solvent.

but don't we have 1163g of solution?

 

[tranv117]

For #1, this is how they started.
0.1m = (x/315.34)/ [(500-x)/1000]
confusing... 🙁



but don't we have 1163g of solution?

again, its looking at molarity and molality. molal= mol solute/kg solvent. x/MW will give you moles of X solute. The question gives you 500g of aqueous soln. So you cant just divide by .5kg. You want the weight of the solvent itself so you have to take 500 minus the weight of the solute to give you the weight of just the solvent.

 

[tranv117]

For #1, this is how they started.
0.1m = (x/315.34)/ [(500-x)/1000]
confusing... 🙁



but don't we have 1163g of solution?

No. that is the density. which is 1163g per liter of solution. You have 1180 grams of solution so dividing 1180/1163 gives you L of solution.

 

[tranv117]

yeah, these questions are a bit tricky. But if you get Molarity and molality straight, Its really a matter of multiplying/dividing and crossing units out.

 

[joonkimdds]

Here is what I am really confused.
#35 destroyer says
"A solution of LiCl has a density of 1.13g/cm^3. If the solution is 20% by weight LiCl, which is the molarity?"
what they did was they assumed that they have 1130g of solution and 1L.
They got this info from the density.
From here they got 20% of 1130g and converted into mol which is 5.3mol.
and since they got 1L of solution from the density, they have M = 5.3mol/1L = 5.3M.

This question makes me think that I get the mass of solution and L of solution from the density.


and now if u look at the following question
"What is the molarity of a 3.21 m KOH(aq) solution? (density of solution = 1.163 g/mL)"
they gave us density.
So what did I think first? from the DAT destroyer lesson, I learned that I should get the mass and L of solution from the density.
That's why I said I must have 1163g and 1L of solution.

 

[anesthesia_209]

How many grams of BiCl3 are needed to make 500.0 g of a 0.10-m aqueous solution?
I don’t know how to do this.
The answer is 15.28g

For this one, I figured out how many grams are in .1mol. (31.5), then added it to 1L solvent (since its water, 1L = 1kg). This gives you 1031.5. They want 500g of this so I just turn it into a percentage (500g/1031.5g). Then I just multiply that by the number of grams in the 1kg of solvent (31.5) to give you 15.28.

 

[utdent20]

i am totally lost on this problem can someone clearly put the steps to solving the #2 problem.

 

[Tina324]

yea can someone re-explain #2 paleeease

 

[Mamona]

#2 please nice explanation!!!!😀

 

[anesthesia_209]

3.21mol KOH = 180.081g
1L H20 = 1kg

Add them together = 1180.081g
Now use density to find out how many liters it is.

1180.081g/1.163 = 1014.69mL = 1.015L

Now use the number of moles of KOH and find out the molarity.

3.21mol KOH/1.015L = 3.16M