Chemistry Help Forum concentration and molarity problems

 Nov 30th 2011, 11:35 AM #1 nesperson Junior Member   Join Date: Nov 2011 Posts: 3 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)
 Nov 30th 2011, 10:44 PM #2 Unknown008 Senior Member     Join Date: May 2010 Location: Mauritius Posts: 1,685 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? __________________ Jerry (Working, studying, writing and drawing :3) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying? Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime. Thank you for leaving a reputation to my post
 Dec 1st 2011, 10:24 AM #3 nesperson Junior Member   Join Date: Nov 2011 Posts: 3 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...
Dec 1st 2011, 10:25 AM   #4
nesperson
Junior Member

Join Date: Nov 2011
Posts: 3
 Originally Posted by Unknown008 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...

 Dec 1st 2011, 10:27 AM #5 Unknown008 Senior Member     Join Date: May 2010 Location: Mauritius Posts: 1,685 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. __________________ Jerry (Working, studying, writing and drawing :3) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying? Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime. Thank you for leaving a reputation to my post
Dec 2nd 2011, 01:36 PM   #6
bjhopper
Senior Member

Join Date: May 2010
Posts: 491
 Originally Posted by nesperson 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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