Download presentation

Presentation is loading. Please wait.

Published byFrancis Hardy Modified over 6 years ago

1
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

2
#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

3
#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

4
#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

5
#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

6
#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

7
#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

8
#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

9
#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

10
#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

11
#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)

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google