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Topic: Density and diluted solutions  (Read 15633 times)

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Offline student8607

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Density and diluted solutions
« on: November 06, 2008, 06:53:47 PM »
I'm very comfortable with working with typical diluted solutions problems using

MiVi = MfVf
initial molarity and volume = final molarity and volume

I can figure out the molarity using (moles)/(L) BUT
how would that work with densities?


EX: The concentration of Mg2+ seawater is 1.29g/kg. If the density of seawater is 1.04g/ml, what is the molarity of Mg2+?

Offline DrCMS

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Re: Density and diluted solutions
« Reply #1 on: November 06, 2008, 07:22:55 PM »
I'm very comfortable with working with typical diluted solutions problems using

MiVi = MfVf
initial molarity and volume = final molarity and volume

I can figure out the molarity using (moles)/(L) BUT
how would that work with densities?

How do you think it will work with density?  What is the definition of molarity?  What are the units for density.  What are the units for the concentration in the question?  Now put them all together.

Offline student8607

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Re: Density and diluted solutions
« Reply #2 on: November 06, 2008, 07:39:03 PM »
D = grams (mass) / mL (volume)
M = mole (mass) / L (volume)
concentration = g/kg....WAIT: I thought molarity WAS concentration?

If I can get kg to L  (maybe using density somehow?) and then g to moles, I would have concentration for Mg


Offline DrCMS

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Re: Density and diluted solutions
« Reply #3 on: November 06, 2008, 07:54:03 PM »
D = grams (mass) / mL (volume)
Yes density is mass per unit volume.

M = mole (mass) / L (volume)
Molarity is moles per L.

concentration = g/kg....WAIT: I thought molarity WAS concentration?
No molarity is just one measure of concentration, g/kg is another, ppm is another, % is another, molality is another etc. etc.

If I can get kg to L  (maybe using density somehow?) and then g to moles, I would have concentration for Mg
Yes

Offline student8607

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Re: Density and diluted solutions
« Reply #4 on: November 06, 2008, 07:58:52 PM »
If I divide:
(1.29g/kg) / (1.04g/mL) = kgmL

If I multiple: g^2kg/mL

How can I get to just mL?

Offline DrCMS

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Re: Density and diluted solutions
« Reply #5 on: November 06, 2008, 08:18:29 PM »
If I divide:
(1.29g/kg) / (1.04g/mL) = kgmL

If I multiple: g^2kg/mL

How can I get to just mL?

1.04g/ml means 1ml of the sea water weighs 1.04g
So what does 1L weigh?

There are 1.29g of Mg2+ in 1kg of sea water
How many  grams of Mg2+ ions are in that 1L of sea water?

How many moles is that?

Offline student8607

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Re: Density and diluted solutions
« Reply #6 on: November 06, 2008, 08:32:41 PM »
if 1 mL = 1.04g of seawater
then 1L = 1.04x10^-3g of seawater

(1.29g) / (1.04x10^-3g/L) = 1240L

1.29gMg x 1mol / 24.305 = 0.0531mol Mg

0.0531mol Mg / 1240L Mg = 4.28x10^-5M ?

Offline DrCMS

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Re: Density and diluted solutions
« Reply #7 on: November 06, 2008, 08:43:48 PM »
You are just not listening are you. 
I showed you exactly how to do it step by step and you ignored it.

Does is really make sense that 1ml weighs 1.04g but 1L weighs 1.04mg?

How many ml in 1L?

Now start again with the steps i told you to follow:

1.04g/ml means 1ml of the sea water weighs 1.04g
So what does 1L weigh?

There are 1.29g of Mg2+ in 1kg of sea water
How many  grams of Mg2+ ions are in that 1L of sea water?

How many moles is that?

Offline student8607

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Re: Density and diluted solutions
« Reply #8 on: November 06, 2008, 08:50:09 PM »
I'm SORRY. I'm trying my best.

0.001mL in 1L
1.04g/ml means 1ml of the sea water weighs 1.04g
So 1L of sea water weighs 0.00104g

There are 1.29g of Mg2+ in 1kg of sea water
How many  grams of Mg2+ ions are in that 1L of sea water?
I DONT KNOW HOW YOU CAN CONVERT KG TO L?

Offline DrCMS

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Re: Density and diluted solutions
« Reply #9 on: November 06, 2008, 08:54:14 PM »
No 1000ml = 1L

Offline student8607

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Re: Density and diluted solutions
« Reply #10 on: November 06, 2008, 09:00:23 PM »
My mistake.
Still don't see how this is going to come together.

Offline DrCMS

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Re: Density and diluted solutions
« Reply #11 on: November 06, 2008, 09:23:15 PM »
Work through the steps i gave you

1.04g/ml means 1ml of the sea water weighs 1.04g
So what does 1L weigh?

If there are 1.29g of Mg2+ in 1kg of sea water how many grams of Mg2+ ions are in 1L of sea water?

How many moles is that?

Giving you a final molarity answer as moles per litre

Offline student8607

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Re: Density and diluted solutions
« Reply #12 on: November 06, 2008, 09:48:04 PM »
1.04g/ml means 1ml of the sea water weighs 1.04g
So 1L of sea water weighs 0.00104g
(mL is smaller than liter)

If there are 1.29g of Mg2+ in 1kg of sea water how many grams of Mg2+ ions are in 1L of sea water?
0.00104g --> 1.04kg sea water
1.29g of Mg in 1.04kg OR L sea water

1.29/1.04 ?

Offline DrCMS

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Re: Density and diluted solutions
« Reply #13 on: November 06, 2008, 10:01:00 PM »
1.04g/ml means 1ml of the sea water weighs 1.04g
So 1L of sea water weighs 0.00104g
(mL is smaller than liter)

NO NO NO!  How many times do you need to be told something before you take it in.
If 1ml is smaller than 1L why do you think 1L weighs less than 1ml????
1L = 1000ml
so if 1ml = 1.04g
1L = 1000x1.04g = 1040g = 1.04kg

1.29g of Mg in 1.04kg

NO NO NO! The original question tells you there is 1.29g Mg in 1kg.

Given that info how many grams of Mg in 1L where 1L = 1.04kg?

Offline enahs

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Re: Density and diluted solutions
« Reply #14 on: November 06, 2008, 10:22:31 PM »
There are 1000mL per 1 L.
Mathematically.

1000mL

   1L

If there are 1000mL in 1 L, then 1 mL is 1000th of a L.
Using dimensional analysis.

1 mL  *     1 L      = 0.001 L = 1 (10-3) L = 1 mL
              1000 mL

So, 425.6789 mL =

425.6789 mL  *   1 L          =.4256789 L
                          1000 mL

If there is 1 g in 1 mL, and 1000 mL in 1 L, then there are 1000g in a L, correct?
In this question you do not have 1g in 1 mL though.

Now, re-read all of DrCMS's posts.



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