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Published byFrancis Hardy Modified over 9 years ago

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Dilution Practice Remember! M 1 V 1 = M 2 V 2 M 1 = concentration of stock solution V 1 = volume of stock solution M 2 = goal concentration V 2 = goal volume

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#1 With a 5.0M stock solution of NH 3, you need to make 0.50L of 1.0 M solution. How much stock solution do you need? M 1 = 5.0 M V 1 = ?? M 2 = 1.0M V 2 = 0.50L V 1 = 0.1L = 100 mL How much distilled water will you add to arrive at your desired solution? 500 mL – 100 mL stock solution = 400 mL water added

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#2 With a 6.0M stock solution of NaOH, you need to make 0.250L of 1.0 M solution. How much stock solution do you need? M 1 = 6.0 M V 1 = ?? M 2 = 1.0M V 2 = 0.250L V 1 = 0.0417L = 41.7 mL How much distilled water will you add to arrive at your desired solution? 250 mL – 41.7 mL stock solution = 208.3 mL water added

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#3 With 100 mL of a 6.0M stock solution, how much 0.5M dilute solution can you make? M 1 = 6.0 M V 1 = 100mL = 0.1L M 2 = 0.5M V 2 = ?? V 2 = 1.2L

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#4 If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution. M 1 = 6.0 M V 1 = 175mL = 0.175 L M 2 = ? V 2 = 1.0L M 2 = 0.28 M

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#5 You need to make 10.0 L of 1.2 M KNO 3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it? M 1 = ? V 1 = 2.5 L M 2 = 1.2 M V 2 = 10.0 L M 1 = 4.8 M

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#6 If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be? M 1 = 0.15 M V 1 = 125 mL = 0.125L M 2 = ? V 2 = 25 mL + 125 mL = 150 mL total = 0.150L M 2 = 0.125 M

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#7 If I add water to 100 mL of a 0.15 M NaOH solution until the final volume is 150 mL, what will the molarity of the diluted solution be? M 1 = 0.15 M V 1 = 100 mL = 0.100L M 2 = ? V 2 = 150 mL = 0.150L M 2 = 0.1 M

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#8 I have 345 mL of a 1.50 M NaCl solution. If I boil the water until the volume of the solution is 250. mL, what will the molarity of the solution be? M 1 = 1.50 M V 1 = 345 mL = 0.345L M 2 = ? V 2 = 250. mL = 0.250L M 2 = 2.07 M

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#9 How much water would I need to add to 500. mL of a 2.4 M KCl solution to make a 1.0 M solution? M 1 = 2.4 M V 1 = 500 mL = 0.500L M 2 = 1.0M V 2 = ? V 2 = 1.2 L total, this means you would need to add 0.7 L of pure water to your original solution

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#10 Draw what happens when calcium nitrate dissolves in water. What pieces does it separate into? In what ratio? Ca +2 (aq), 2 NO 3 - (aq)

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