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HW - Concentrations of solutions · PDF fileHonors Chemistry Name _____ Concentrations of ......

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1 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 ! 1 1 ! 169.872 ! = 0.0588678 ! = ( ) = 0.0588678 ! 0.75 = 0.078 ! 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 1 1 74.551 = 0.067068 = βˆ’ = = 0.067068 0.25 = 0.27 = 270 3. What is the molality of a solution that contains 48 grams of sodium chloride and 250 mL of water? 48 1 1 58.443 = 0.8213 250 ! 1 1 ! 1 ! 1 ! 1000 ! = 0.25 ! = = 0.8213 0.25 ! = 3.3 4. What is the percentage by mass of the solution from problem 1? 750 ! 1 1 ! 1 ! = 750 ! = 100 = 10.0 ! 10.0 + 750 100 = 1.32 ! 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? = 100
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Page 1: HW - Concentrations of solutions · PDF fileHonors Chemistry Name _____ Concentrations of ... What is the molarity of a solution that contains 10.0 grams of Silver ... Concentrations

  1  

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  π‘”  π΄π‘”𝑁𝑂!

1  π‘₯  

1  π‘šπ‘œπ‘™π‘’  π΄π‘”𝑁𝑂!169.872  π‘”  π΄π‘”𝑁𝑂!

=  0.0588678  π‘šπ‘œπ‘™π‘’𝑠  π΄π‘”𝑁𝑂!

𝑀 =  π‘›

𝑉  (𝑖𝑛  πΏ)  =  

0.0588678  π‘šπ‘œπ‘™π‘’𝑠  π΄π‘”𝑁𝑂!0.75  πΏ

= 0.078  π‘€  π΄π‘”𝑁𝑂!

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  π‘”  πΎπΆπ‘™1

 π‘₯  1  π‘šπ‘œπ‘™π‘’  πΎπΆπ‘™74.551  π‘”  πΎπΆπ‘™

= 0.067068  π‘šπ‘œπ‘™π‘’𝑠  πΎπΆπ‘™

 π‘€ =  

π‘›π‘‰βˆ’ π‘ π‘œπ‘™π‘£π‘’  π‘“π‘œπ‘Ÿ  π‘£π‘œπ‘™π‘’π‘šπ‘’

𝑉 =  π‘›π‘€=  0.067068  π‘šπ‘œπ‘™π‘’𝑠  πΎπΆπ‘™

0.25  π‘€  πΎπΆπ‘™= 0.27  πΏ = 270  π‘šπΏ

3. What is the molality of a solution that contains 48 grams of sodium chloride and 250 mL of

water?

48  π‘”  π‘π‘ŽπΆπ‘™1

 π‘₯  1  π‘šπ‘œπ‘™π‘’  π‘π‘ŽπΆπ‘™58.443  π‘”  π‘π‘ŽπΆπ‘™

=  0.8213  π‘šπ‘œπ‘™π‘’𝑠  π‘π‘ŽπΆπ‘™

250  π‘šπΏ  π»!𝑂

1  π‘₯  

1  π‘”  π»!𝑂1  π‘šπΏ  π»!𝑂

 π‘₯  1  π‘˜π‘”  π»!𝑂1000  π‘”  π»!𝑂

=  0.25  π‘˜π‘”  π»!𝑂

π‘šπ‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ =  π‘šπ‘œπ‘™π‘’𝑠  π‘ π‘œπ‘™π‘’π‘‘π‘’π‘˜π‘”  π‘ π‘œπ‘™π‘£π‘’𝑛𝑑

=  0.8213  π‘šπ‘œπ‘™π‘’𝑠  π‘π‘ŽπΆπ‘™

0.25  π‘˜π‘”  π»!𝑂= 3.3  π‘š  π‘π‘ŽπΆπ‘™

4. What is the percentage by mass of the solution from problem 1?

750  π‘šπΏ  π»!𝑂

1  π‘₯  

1  π‘”  π»!𝑂1  π‘šπΏ  π»!𝑂

=  750  π‘”  π»!𝑂

π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘  π‘π‘¦  π‘€π‘Žπ‘ π‘  =  π‘šπ‘Žπ‘ π‘   π‘œπ‘“  π‘ π‘œπ‘™π‘’π‘‘π‘’π‘šπ‘Žπ‘ π‘   π‘œπ‘“  π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›

 π‘₯  100 =  10.0  π‘”  π΄π‘”𝑁𝑂!10.0  π‘” + 750  π‘”

 π‘‹  100 = 1.32    π΄π‘”𝑁𝑂!

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?

π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘  π‘‰π‘œπ‘™π‘’π‘šπ‘’ =  π‘£π‘œπ‘™π‘’π‘šπ‘’  π‘œπ‘“  π‘ π‘œπ‘™π‘’π‘‘π‘’π‘£π‘œπ‘™π‘’π‘šπ‘’  π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›

 π‘₯  100

Page 2: HW - Concentrations of solutions · PDF fileHonors Chemistry Name _____ Concentrations of ... What is the molarity of a solution that contains 10.0 grams of Silver ... Concentrations

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π‘‰π‘œπ‘™π‘’π‘šπ‘’  π‘†π‘œπ‘™π‘’𝑑𝑒 =  π‘ƒπ‘’π‘Ÿπ‘π‘’π‘›π‘‘  π‘‰π‘œπ‘™π‘’π‘šπ‘’  π‘₯  π‘‰π‘œπ‘™π‘’π‘šπ‘’  π‘†π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›

100=  (8.5%)(450  π‘šπΏ)

100= 38  π‘šπΏ  π»!𝑂!

6. What is the mole fraction of the solute in the solution from problem 1?

750  π‘šπΏ  π»!𝑂

1  π‘₯  

1  π‘”  π»!𝑂1  π‘šπΏ  π»!𝑂

 π‘₯  1  π‘šπ‘œπ‘™π‘’  π»!𝑂18.015  π‘”  π»!𝑂

= 41.63197  π‘šπ‘œπ‘™π‘’  π»!𝑂

𝑋!"#$%& =  π‘šπ‘œπ‘™π‘’𝑠  π‘ π‘œπ‘™π‘’π‘‘π‘’π‘šπ‘œπ‘™π‘’π‘   π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›

=  0.0588678  π‘šπ‘œπ‘™π‘’𝑠  π΄π‘”𝑁𝑂!

0.0588678  π‘šπ‘œπ‘™π‘’𝑠  π΄π‘”𝑁𝑂! + 41.63197  π‘šπ‘œπ‘™π‘’  π»!𝑂= 0.0014

7. What is the mole fraction of the solvent in the solution from problem 1?

𝑋!"#$%&' =  π‘šπ‘œπ‘™π‘’𝑠  π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘π‘šπ‘œπ‘™π‘’π‘   π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›

=  41.63197  π‘šπ‘œπ‘™π‘’  π»!𝑂

0.0588678  π‘šπ‘œπ‘™π‘’𝑠  π΄π‘”𝑁𝑂! + 41.63197  π‘šπ‘œπ‘™π‘’  π»!𝑂= 1.0

8. What is the molality of the ions in the solution from problem 3?

3.3  π‘š  π‘π‘ŽπΆπ‘™ =  0.82131  π‘šπ‘œπ‘™π‘’𝑠  π‘π‘ŽπΆπ‘™

0.25  π‘˜π‘”  π»!𝑂

1  π‘šπ‘œπ‘™π‘’  π‘π‘ŽπΆπ‘™ = 2  π‘šπ‘œπ‘™π‘’𝑠  π‘–π‘œπ‘›π‘ 

0.82131  π‘šπ‘œπ‘™π‘’𝑠  π‘π‘ŽπΆπ‘™

0.25  π‘˜π‘”  π»!𝑂  π‘₯  2  π‘šπ‘œπ‘™π‘’𝑠  π‘π‘ŽπΆπ‘™  π‘–π‘œπ‘›π‘ 

1  π‘šπ‘œπ‘™π‘’  π‘π‘ŽπΆπ‘™= 6.6  π‘š  π‘π‘ŽπΆπ‘™  π‘–π‘œπ‘›π‘ 

9. What is the molality of a solution that contains 13.4 grams of calcium chloride dissolved in 655

mL of water?

13.4  π‘”  πΆπ‘ŽπΆπ‘™!1

 π‘₯  1  π‘šπ‘œπ‘™π‘’  πΆπ‘ŽπΆπ‘™!

110.986  π‘”  πΆπ‘ŽπΆπ‘™!=  0.1207359  π‘šπ‘œπ‘™π‘’𝑠  πΆπ‘ŽπΆπ‘™!

655  π‘šπΏ  π»!𝑂

1  π‘₯  

1  π‘”  π»!𝑂1  π‘šπΏ  π»!𝑂

 π‘₯  1  π‘˜π‘”  π»!𝑂1000  π‘”  π»!𝑂

=  0.655  π‘˜π‘”  π»!𝑂

π‘šπ‘œπ‘™π‘Žπ‘™π‘–π‘‘π‘¦ =  π‘šπ‘œπ‘™π‘’𝑠  π‘ π‘œπ‘™π‘’π‘‘π‘’π‘˜π‘”  π‘ π‘œπ‘™π‘£π‘’𝑛𝑑

=  0.1207359  π‘šπ‘œπ‘™π‘’𝑠  πΆπ‘ŽπΆπ‘™!

0.655  π‘˜π‘”  π»!𝑂= 0.184  π‘š  πΆπ‘ŽπΆπ‘™!

10. What is the molality of the ions in the solution from problem 9?

0.1207359  π‘šπ‘œπ‘™π‘’𝑠  πΆπ‘ŽπΆπ‘™!

0.655  π‘˜π‘”  π»!𝑂= 0.184  π‘š  πΆπ‘ŽπΆπ‘™!

1  π‘šπ‘œπ‘™π‘’  πΆπ‘ŽπΆπ‘™! = 3  π‘šπ‘œπ‘™π‘’𝑠  πΆπ‘ŽπΆπ‘™!  π‘–π‘œπ‘›π‘ 

0.1207359  π‘šπ‘œπ‘™π‘’𝑠  πΆπ‘ŽπΆπ‘™!

0.655  π‘˜π‘”  π»!𝑂  π‘₯  3  π‘šπ‘œπ‘™π‘’𝑠  πΆπ‘ŽπΆπ‘™!  π‘–π‘œπ‘›π‘ 

1  π‘šπ‘œπ‘™π‘’  πΆπ‘ŽπΆπ‘™!=  0.553  π‘š  πΆπ‘ŽπΆπ‘™!  π‘–π‘œπ‘›π‘ 


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