honors chemistry name · ! 1! honors chemistry name _____ concentrations of solutions date _____...
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Honors Chemistry Name _______________________ Concentrations of Solutions Date ________________________ Complete the following problems on a separate sheet of paper. Use significant figures.
Note: The density of water is 1 g/mL.
1. What is the molarity of a solution that contains 10.0 grams of Silver Nitrate that has been dissolved in 750 mL of water?
10.0 𝑔 𝐴𝑔𝑁𝑂!
1 𝑥
1 𝑚𝑜𝑙𝑒 𝐴𝑔𝑁𝑂!169.872 𝑔 𝐴𝑔𝑁𝑂!
= 0.0588678 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂!
𝑀 = 𝑛
𝑉 (𝑖𝑛 𝐿) =
0.0588678 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂!0.75 𝐿
= 0.078 𝑀 𝐴𝑔𝑁𝑂!
2. You want to create a 0.25 M Potassium Chloride solution. You mass 5.00 grams of Potassium
Chloride. How much water is needed?
5.00 𝑔 𝐾𝐶𝑙1
𝑥 1 𝑚𝑜𝑙𝑒 𝐾𝐶𝑙74.551 𝑔 𝐾𝐶𝑙
= 0.067068 𝑚𝑜𝑙𝑒𝑠 𝐾𝐶𝑙
𝑀 =
𝑛𝑉− 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
𝑉 = 𝑛𝑀= 0.067068 𝑚𝑜𝑙𝑒𝑠 𝐾𝐶𝑙
0.25 𝑀 𝐾𝐶𝑙= 0.27 𝐿 = 270 𝑚𝐿
3. What is the molality of a solution that contains 48 grams of sodium chloride and 250 mL of
water?
48 𝑔 𝑁𝑎𝐶𝑙1
𝑥 1 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶𝑙58.443 𝑔 𝑁𝑎𝐶𝑙
= 0.8213 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
250 𝑚𝐿 𝐻!𝑂
1 𝑥
1 𝑔 𝐻!𝑂1 𝑚𝐿 𝐻!𝑂
𝑥 1 𝑘𝑔 𝐻!𝑂1000 𝑔 𝐻!𝑂
= 0.25 𝑘𝑔 𝐻!𝑂
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 = 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
= 0.8213 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
0.25 𝑘𝑔 𝐻!𝑂= 3.3 𝑚 𝑁𝑎𝐶𝑙
4. What is the percentage by mass of the solution from problem 1?
750 𝑚𝐿 𝐻!𝑂
1 𝑥
1 𝑔 𝐻!𝑂1 𝑚𝐿 𝐻!𝑂
= 750 𝑔 𝐻!𝑂
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑏𝑦 𝑀𝑎𝑠𝑠 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 100 = 10.0 𝑔 𝐴𝑔𝑁𝑂!10.0 𝑔 + 750 𝑔
𝑋 100 = 1.32 𝐴𝑔𝑁𝑂!
5. How many mL of hydrogen peroxide are needed to make a 8.5% solution by volume of hydrogen
peroxide if you want to make 450 mL of solution?
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒𝑣𝑜𝑙𝑢𝑚𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑥 100
2
𝑉𝑜𝑙𝑢𝑚𝑒 𝑆𝑜𝑙𝑢𝑡𝑒 = 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑉𝑜𝑙𝑢𝑚𝑒 𝑥 𝑉𝑜𝑙𝑢𝑚𝑒 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
100= (8.5%)(450 𝑚𝐿)
100= 38 𝑚𝐿 𝐻!𝑂!
6. What is the mole fraction of the solute in the solution from problem 1?
750 𝑚𝐿 𝐻!𝑂
1 𝑥
1 𝑔 𝐻!𝑂1 𝑚𝐿 𝐻!𝑂
𝑥 1 𝑚𝑜𝑙𝑒 𝐻!𝑂18.015 𝑔 𝐻!𝑂
= 41.63197 𝑚𝑜𝑙𝑒 𝐻!𝑂
𝑋!"#$%& = 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 0.0588678 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂!
0.0588678 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂! + 41.63197 𝑚𝑜𝑙𝑒 𝐻!𝑂= 0.0014
7. What is the mole fraction of the solvent in the solution from problem 1?
𝑋!"#$%&' = 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑣𝑒𝑛𝑡𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 41.63197 𝑚𝑜𝑙𝑒 𝐻!𝑂
0.0588678 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂! + 41.63197 𝑚𝑜𝑙𝑒 𝐻!𝑂= 1.0
8. What is the molality of the ions in the solution from problem 3?
3.3 𝑚 𝑁𝑎𝐶𝑙 = 0.82131 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
0.25 𝑘𝑔 𝐻!𝑂
1 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶𝑙 = 2 𝑚𝑜𝑙𝑒𝑠 𝑖𝑜𝑛𝑠
0.82131 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙
0.25 𝑘𝑔 𝐻!𝑂 𝑥 2 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶𝑙 𝑖𝑜𝑛𝑠
1 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶𝑙= 6.6 𝑚 𝑁𝑎𝐶𝑙 𝑖𝑜𝑛𝑠
9. What is the molality of a solution that contains 13.4 grams of calcium chloride dissolved in 655
mL of water?
13.4 𝑔 𝐶𝑎𝐶𝑙!1
𝑥 1 𝑚𝑜𝑙𝑒 𝐶𝑎𝐶𝑙!
110.986 𝑔 𝐶𝑎𝐶𝑙!= 0.1207359 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑙!
655 𝑚𝐿 𝐻!𝑂
1 𝑥
1 𝑔 𝐻!𝑂1 𝑚𝐿 𝐻!𝑂
𝑥 1 𝑘𝑔 𝐻!𝑂1000 𝑔 𝐻!𝑂
= 0.655 𝑘𝑔 𝐻!𝑂
𝑚𝑜𝑙𝑎𝑙𝑖𝑡𝑦 = 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
= 0.1207359 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑙!
0.655 𝑘𝑔 𝐻!𝑂= 0.184 𝑚 𝐶𝑎𝐶𝑙!
10. What is the molality of the ions in the solution from problem 9?
0.1207359 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑙!
0.655 𝑘𝑔 𝐻!𝑂= 0.184 𝑚 𝐶𝑎𝐶𝑙!
1 𝑚𝑜𝑙𝑒 𝐶𝑎𝐶𝑙! = 3 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑙! 𝑖𝑜𝑛𝑠
0.1207359 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑙!
0.655 𝑘𝑔 𝐻!𝑂 𝑥 3 𝑚𝑜𝑙𝑒𝑠 𝐶𝑎𝐶𝑙! 𝑖𝑜𝑛𝑠
1 𝑚𝑜𝑙𝑒 𝐶𝑎𝐶𝑙!= 0.553 𝑚 𝐶𝑎𝐶𝑙! 𝑖𝑜𝑛𝑠