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Solutions Class #6

Solutions Class #6. Partner up again, different partners today please. According to an article in the New England Journal of Medicine, Vol. 349, October 30, 2003, mercury toxicity begins at 0.100 PPM.

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Solutions Class #6

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  1. Solutions Class #6 Partner up again, different partners today please.

  2. According to an article in the New England Journal of Medicine, Vol. 349, October 30, 2003, mercury toxicity begins at 0.100 PPM. If a nutjob dropped one pound of mercury into a school pool, with a volume of 1,129,000 liters, would you be able to safely swim in there? (1 pound = 454 grams)

  3. According to an article in the New England Journal of Medicine, Vol. 349, October 30, 2003, mercury toxicity begins at 0.100 PPM. If a nut dropped one pound of mercury into a school pool, with a volume of 1,129,000 liters, would you be able to safely swim in there? Short answer: NO. PPM = x 1,000,000 PPM = x 1,000,000 PPM = x 1,000,000 PPM = 0.402 PPM this is more than 4x the safe levels of mercury grams Hggrams solution 454 g Hg 1,129,000,000 g 454 g Hg 1,129,000,000 g

  4. How many parts per million is NaCl in a 1.00 M salt water solution of 1.00 liters?

  5. How many parts per million is NaCl in a 1.00 M salt water solution of 1.00 liters? Figure out how many moles NaCl are present, 1 molar means one mole per liter, so one mole NaCl is 58 grams of salt. The solution is 1000 mL, or 1000 grams grams solutegrams solution PPM = x 1,000,000 58 g NaCl1000 g solution PPM = x 1,000,000 PPM = 58,000 parts per million this is silly of course, molarity is a more sensible measure here

  6. What is the molarity of a solution where 148 g KCl is dissolved into a solution of 5000. mL total volume?

  7. What is the molarity of a solution where 148 g KCl is dissolved into a solution of 5000. mL total volume? moles of soluteliters of solution M = M = = 0.400 molar (3sf) 2.00 moles KCl5.000 liters

  8. How do you mix up a 25.5 mL solution that is 0.850 Molar NaOH(AQ), if you use a stock solution of 6.40 molarity. Then draw a diagram to show how to “make” this solution.

  9. How do you mix up a 25.5 mL solution that is 0.850 Molar NaOH(AQ), if you use a stock solution of 6.40 molarity. Then draw a diagram to show how to “make” this solution. M1V1 = M2V2 (6.40 M)(V1) = (O.850 M)(25.5 mL) V1 = 3.39 mL stock (3 sf) Next slide please… this is NOT it.

  10. 25.5 mL total volume – 3.39 mL stock = 22.1 mL water for the dilution (3 SF) That’s the 25.5 mL fill up to here line SECOND: fill to 25.5 mL total volume (with water) FIRST: put 3.39 mL stock solution into beaker.

  11. You dissolve 2.25 moles of potassium bromide into water forming a 1.00 liter solution. What is this solution’s boiling point, and freezing point, IN KELVIN? (don’t worry about the SF in this problem, please)

  12. You dissolve 2.25 moles of potassium bromide into water forming a 2.00 liter solution. What is this solution’s boiling point, and freezing point? (No SF here) 2.25 moles KBr yields 2.25 moles K+1 and 2.25 moles of Br-1 = 4.5 moles of ions total New Boiling point of solution 373 Kelvin + (4.5 x 0.50 Kelvin) = 373 Kelvin + 2.25 Kelvin = 375.25 Kelvin (SF be gone!) New Freezing point of solution 273 Kelvin - (4.5 x 1.86 Kelvin) = 273 Kelvin – 8.37 K = 264.63 Kelvin (SF don’t count here)

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  14. You have two 1.0 liter glasses of solution of equal volume in the same room (same temp and pressure). One solution is a 3.50 M NaCl aqueous solution, the other is a 2.00 M Ca(NO3)2(AQ) Which one would evaporate dry first, and why? You must “prove” your answer with some math NaCl(AQ) Ca(NO3)2(AQ)

  15. You have 2 beakers of solution of equal volume in the same room (same temp and pressure). One solution is a 3.50 M NaCl aqueous solution, the other is a 2.00 M Ca(NO3)2(AQ) The sodium chloride solution has 3.50 moles x 2 ions per mole, or 7 moles of ions in solution. The calcium nitrate solution has 2.00 moles x 3 ions per mole, or 6 moles of ions in solution. The solution with the MOST moles of ions in solution will evaporate more slowly.It will have a lower vapor pressure. NaCl(AQ) Ca(NO3)2(AQ)

  16. If you have a 2.40 M HCl stock solution, how do you make 50.0 mL of 3.00 M HCl from it? A diagram would help you think through this math.

  17. If you have a 2.40 M HCl stock solution, how do you make 50.0 mL of 3.00 M HCl from it? A diagram would help you think through this math. Nothing would really help, no solutions can be made stronger than your stock solution. Silly rabbit.

  18. What volume of solution contains 475 g of sodium chloride at 0.933 Molarity?

  19. What volume of solution contains 475 g of sodium chloride at 0.933 Molarity? # Moles Liters of solution Molarity = 8.19 Moles X Liters 0.933 M = 1 0.933 (X) = 8.19 X = 8.78 liters total volume

  20. In front of you are 3 solutions labeled. Which has the highest boiling point, which has the lowest freezing point? 1.50 M CaCl2(AQ) 3.00 M NCl3(AQ) 2.00 M NaCl(AQ)

  21. In front of you are 3 solutions labeled. Which has the highest boiling point, which has the lowest freezing point? 1.50 M CaCl2(AQ) 4.5 moles of particles Highest BP Lowest FP 3.00 M NCl3(AQ) 3 moles of particles 2.00 M NaCl(AQ) 4 moles of particles

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