Today is 26 March 2010 -- Solution Chemistry Exam

This exam covers solution chemistry.  Partial answers are provided.  You will need to follow the logic and complete the problems for credit.   Remember that "(aq)" means aqueous (e.g. -- "in water"). 

i.  "Define Molarity"

YOU Check the text definition.

ii. Define "Molality"

YOU Check the text definition.

iii.  Define "parts per million."

YOU Check the text definition.

1. A 0.750 L aqueous solution contains 90.0 g of ethanol, C2H5OH. Calculate the molar concentration of the solution in mol/Liter

Want molarity (M) = moles/liter.  Moles = grams/atomic wt.  So M = moles/0.75 liter

2. What mass of NaCl is dissolved in 152 mL of a solution if the concentration of the solution is 0.364 M?

Want mass NaCl.  M = moles/liter = 0.364 molar = moles/0.152 Liter.  Solve for moles.

Then, since moles = mass/At. Wt. Rearrange for (At. Wt.)(moles) = mass

3. What mass of dextrose, C6H12O6 is dissolved in 325 mL of 0.258 M solution?

Want mass  C6H12O6.  M = moles/liter  0.248 M = moles/0.325 liter

mass = moles x at. wt.

4. A mass of 98 g of sulfuric acid, H2SO4, is dissolved in water to prepare a 0.500 M solution. What is the volume of the solution?

Want Volume.  Find moles; moles = mass H2SO4/At. Wt.  Then Volume = moles/0.5 M

5. A solution of sodium carbonate, Na2CO3, contains 53.0 g of solute in 215 mL of solution. What is its molarity?

Want molarity.  Find moles; moles = 53 g Na2CO3/At. Wt Na2CO3. 

Then M = moles/0.215 Liter

6. What is the molarity of a solution of HNO3 that contains 12.6 g of solute in 5.00 L of solution?

Want molarity.  Find moles; moles = 12.6 g HNO3/At. Wt. HNO3.  Then M = moles/5.0 liter.

7. What mass of copper(II) nitrate, Cu(NO3)2, is present in 50.00 mL of a 4.55 x 10-3 M aqueous solution?

Want mass.  M = moles/liter; 4.55 x 10-3 M = moles/0.050 liter.  Then mass = moles x at. wt. of Cu(NO3)2

8.  What volume of 0.778 M Na2CO3 (aq) solution should be diluted to 150.0 ml with water to reuduce its concentration to 0.0234 M Na2CO3 (aq)?

Want volume.  First find moles: M = moles/Liter.  Rearrange for (0.778 M Na2CO3) x 1 liter = moles.  Then divide by 1000 ml/liter for moles per milliliter.  Then calculate how many moles needed for dilute solution; M = moles/liter = 0.0234 M Na2CO3 = moles/0.150 liter.  Since M means "moles per liter," the liter unit cancels out when you do the math.  You have moles needed which you divide by the answer to the first part.  That is in moles per milliliter.  Moles unit cancels and you have answer in milliliters.

9.  A 25 ml sample of HCl (aq) is diluted to a volume of 500 ml.  If the concentration of the diluted solution is 0.085 M, what was the concentration of the original solution?

Want Molarity.  Dil HCl = 0.085 M = moles HCl/liter.  Rearrange to find moles HCl.  Then find how many moles in 0.5 liter:  0.5 liter x moles/liter = moles available.

Original concentration = M = moles available/orig. vol. = concentration, M