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Old Nov 30th 2011, 12:35 PM   #1
nesperson
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concentration and molarity problems

I need help with these problems please!








ch. 9 drills
17. a) a stock bottle of aqueous sulfuric acid indicates that the solution is 90.0% H2SO4 by mass and has a density of 1.2 g/mL. Calculate the molarity of the solution (Ans 11.0 M)

e) If the concentration of Na+ (sodium ion) in an aqueous solution of sodium phosphate is 3.2x10^-1 mol/L, how many grams of sodium phosphate must have been dissolved in 500 mL to make the solution? (Ans 8.75g)

18. What volume of 0.185 M MgCl2 must be added to 235 mL of 0.206 M KCL to produce a solution with a concentration of 0.250 M Cl- (Chlorine ion)? (Ans 86.2 mL)


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Old Nov 30th 2011, 11:44 PM   #2
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17 a)

The easiest way is to take 1 L. The density is 1.2 g/mL, can you get the mass of 1 L of the acid?

Then, 90% of that mass is H2SO4. Find the number of moles of H2SO4 this represents.

So you have such amount of moles of H2SO4 in 1 L of solution, which is your concentration!

e) Get first the chemical formula of sodium phosphate. Then find the number of moles of sodium ions in 500 mL of solution. Using your chemical formula and the number of moles of sodium ions in the solution, find the number of moles of phosphate ions in the solution.

Find the mass of the sodium ions using the number of moles and add to the mass of the phosphorus ions using the mass of phosphate ions and the number of moles of phosphate ions in the 500 mL solution.

18. Find the number of moles of Cl in the 235 mL of 0.206 M of KCl.

Now, you can use the equation of moles:

Moles from KCl + Moles from MgCl2 = Moles in mixture

Using the formula for concentration;

C1V1 + C2V2 = CV

C1 is the concentration of KCl, V1 is 235 mL of KCl, C2 is 0.185, V2 is what you have to find, C is 0.250, V is the sum of V1 and V2.

Can you solve this equation?
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Old Dec 1st 2011, 11:24 AM   #3
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hi, thank you for your clear instructions on these problems. I was able to solve the first two but I'm still having trouble with 18.
I'm a bit confused from the beginning of your instructions. I calculated moles of KCl to be 0.48 and got the same answer for Cl. Then I didn't know how to find moles from MgCl2 since we only have the Molarity, so I couldn't do the next step...
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Old Dec 1st 2011, 11:25 AM   #4
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Originally Posted by Unknown008 View Post
17 a)

The easiest way is to take 1 L. The density is 1.2 g/mL, can you get the mass of 1 L of the acid?

Then, 90% of that mass is H2SO4. Find the number of moles of H2SO4 this represents.

So you have such amount of moles of H2SO4 in 1 L of solution, which is your concentration!

e) Get first the chemical formula of sodium phosphate. Then find the number of moles of sodium ions in 500 mL of solution. Using your chemical formula and the number of moles of sodium ions in the solution, find the number of moles of phosphate ions in the solution.

Find the mass of the sodium ions using the number of moles and add to the mass of the phosphorus ions using the mass of phosphate ions and the number of moles of phosphate ions in the 500 mL solution.

18. Find the number of moles of Cl in the 235 mL of 0.206 M of KCl.

Now, you can use the equation of moles:

Moles from KCl + Moles from MgCl2 = Moles in mixture

Using the formula for concentration;

C1V1 + C2V2 = CV

C1 is the concentration of KCl, V1 is 235 mL of KCl, C2 is 0.185, V2 is what you have to find, C is 0.250, V is the sum of V1 and V2.

Can you solve this equation?
hi, thank you for your clear instructions on these problems. I was able to solve the first two but I'm still having trouble with 18.
I'm a bit confused from the beginning of your instructions. I calculated moles of KCl to be 0.48 and got the same answer for Cl. Then I didn't know how to find moles from MgCl2 since we only have the Molarity, so I couldn't do the next step...
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Old Dec 1st 2011, 11:27 AM   #5
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To get that number of moles you have to use the next part

Use the values that I gave you:

C1V1 + C2V2 = CV

C1 is the concentration of KCl, V1 is 235 mL of KCl, C2 is 0.185, V2 is what you have to find, C is 0.250, V is the sum of V1 and V2.
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Old Dec 2nd 2011, 02:36 PM   #6
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Originally Posted by nesperson View Post
I need help with these problems please!








ch. 9 drills
17. a) a stock bottle of aqueous sulfuric acid indicates that the solution is 90.0% H2SO4 by mass and has a density of 1.2 g/mL. Calculate the molarity of the solution (Ans 11.0 M)

e) If the concentration of Na+ (sodium ion) in an aqueous solution of sodium phosphate is 3.2x10^-1 mol/L, how many grams of sodium phosphate must have been dissolved in 500 mL to make the solution? (Ans 8.75g)

18. What volume of 0.185 M MgCl2 must be added to 235 mL of 0.206 M KCL to produce a solution with a concentration of 0.250 M Cl- (Chlorine ion)? (Ans 86.2 mL)

For no 18 the concentration of the magnesium chloride in the equation to calculate the volume of it to be added to the 235 ml ofKCL is not 0.185 M because ???
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