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Offline akanevsky

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Molar Concentration Question
« on: May 13, 2011, 08:02:04 AM »
Hello,

I have the following question:

If 50.0 mL of a 0.10 M solution of sodium chloride is mixed with 50.0 mL of 0.10 M magnesium chloride, what is the molar concentration of chloride in the resulting solution?

My attempt at solving the problem:

Sodium Chloride is NaCl.
Magnesium Chloride is MgCl2.

M = moles / liters
moles = M * liters
.10 M * (50 x 10^-3 L) = .005 mol sodium chloride
.10 M * (50 x 10^-3 L) = .005 mol magnesium chloride

In 1 mole of NaCl, there is 1/2 moles of Cl. In .005 moles of NaCl, there is .0025 moles of Cl.
In 1 mole of MgCl2, there is 2/3 moles of Cl. In .005 moles of MgCl2, there is .003333 moles of Cl.

Total moles of Cl: .0025 + .003333 = .00583 moles Cl.

M = moles / liters
M = .00583 moles / (100 * 10^-3 L)
M = .0583

That is the answer that I get.

However, the correct answer is 0.15 M.

Where did I make a mistake?


Offline DrCMS

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Re: Molar Concentration Question
« Reply #1 on: May 13, 2011, 09:19:07 AM »
Where did I make a mistake?

Here

In 1 mole of NaCl, there is 1/2 moles of Cl. In .005 moles of NaCl, there is .0025 moles of Cl.
In 1 mole of MgCl2, there is 2/3 moles of Cl. In .005 moles of MgCl2, there is .003333 moles of Cl.

In 1 mole of NaCl, there is 1 mole of Cl.
In 1 mole of MgCl2, there is 2 moles of Cl.

Offline akanevsky

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Re: Molar Concentration Question
« Reply #2 on: May 13, 2011, 10:13:32 AM »
Thank you so much!

Now I get the right answer from this problem!

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