Entry Task: May 1st - 2nd Block #1Collect Entry Tasks Sheets!!

Agenda:

• Discuss Solutions review w.s.– DON’T LOSE IT!!

• Ice cream “lab”

1. What would be the percent concentration of each of the following solutions?

a. 54.0 g of AgNO3 g is dissolved in 128 g of water.

54 g AgNO3

128g water + 54 g AgNO3 = 182 g of solution X 100

= 29.7%

1. What would be the percent concentration of each of the following solutions?

b. 4.22 g of K2CO3 is dissolved in 426 mL of water.

4.22 g K2CO3

426 g water + 4.22 g K2CO3 = 430.22 g of solution X 100

= 0.981%

2. What mass of solute is needed to produce each of the indicated solutions?

a. 500.0 g of a 6.40% NaCl solution.

X g NaCl500g of solution

X 100 = 6.40%

0.064 X 500 = 32.0 g NaCl

2. What mass of solute is needed to produce each of the indicated solutions?

b. 136 g of a 14.2% LiNO3 solution.

X g LiNO3

136 g of solution X 100 = 14.2%

0.142 X 136 = 19.3 g LiNO3

3. Calculate the grams needed of solutes to prepare the indicated volume and concentration of the solutions given?

a. 340. mL of a 1.82 M aluminum nitrate solution. mm= 212.991g

X g Al(NO3)3

0.340 L solution = 1.82 M

0.340 X 1.82 = 0.619 mol Al(NO3)3

0.619 mol Al(NO3)3

1 mol Al(NO3)3

212.991g Al(NO3)3

=132 g Al(NO3)3

3. Calculate the grams needed of solutes to prepare the indicated volume and concentration of the solutions given?

b. 25.0 mL of a 4.26 M potassium cyanide solution. mm= 65g

X g KCN0.025 L solution

= 4.26 M

0.025 X 4.26 0.1065 mol KCN

0.1065 mol KCN

1 mol KCN

65 g KCN

=6.92 g KCN

4. What should the final volume(mL) of each solution be so that the amount of solute dissolved will produce the indicated concentration.

a. 2.86 g of copper(I) carbonate for a 0.640 M solution. mm= 187.089g

0.0153 mol Cu2CO3

X L solution = 0.640 M

0.0153÷0.640 = 0.0239 L

2.86 g Cu2CO3 1 mol Cu2CO3

187.089g Cu2CO3

= 0.0153 mol Cu2CO3

4. What should the final volume(mL) of each solution be so that the amount of solute dissolved will produce the indicated concentration.

b. 12.62 g of calcium hydrogen carbonate for a 1.28 M solution. mm= 161.994g

0.0779 mol Ca(HCO3)2

X L solution = 1.28 M

0.0779 ÷ 1.28 = 0.0608 L

12.62 g Ca(HCO3)21 mol Ca(HCO3)2

161.994g Ca(HCO3)2

= 0.0779 mol Ca(HCO3)2

5. What will be the final concentration of a solution prepared by dissolving the indicated solute in enough water to produce the indicated volume of solution?

a. 15.4 g of strontium acetate filled up to 340. mL.mm= 205.62g

0.0749 mol Sr(C2H3O2)2

0.340 L solution = 0.220 M

15.4 g Sr(C2H3O2)21 mol Sr(C2H3O2)2

205.62g Sr(C2H3O2)2

= 0.0749 mol Sr(C2H3O2)2

5. What will be the final concentration of a solution prepared by dissolving the indicated solute in enough water to produce the indicated volume of solution?

b. 176.2 g of Iron(III) sulfite filled up to 1.42 liters.mm= 351.591g

0.501 mol Fe2(SO3)3

1.42 L solution = 0.359 M

176.2 g Fe2(SO3)31 mol Fe2(SO3)3

351.591g Fe2(SO3)3

= 0.501 mol Fe2(SO3)3

6. What will be the final concentration of the solution indicated that will result from the following dilutions?

M1V1=M2V2

a. 14.0 mL of a 4.20 M Na2CO3 solution is diluted to 86.0 mL.

58.8

(4.20 M)(14.0ml) = ( XM)(86.0ml)

86.6= 67.9M

6. What will be the final concentration of the solution indicated that will result from the following dilutions?

M1V1=M2V2

b. 450. mL of a 1.22 M HCl solution is diluted to 1.26 liters.

549

(1.22 M)(450ml) = ( XM)(1260 ml)

1260= 0.436M

7. To what volume should the indicated solution be diluted to produce a solution of the desired concentration?

M1V1=M2V2

a. 12.0 mL of a 0.64 M KCl solution for a 0.19 M solution.

7.68

(0.64 M)(12.0ml) = ( 0.19M)(X ml)

0.19= 40.4 ml

7. To what volume should the indicated solution be diluted to produce a solution of the desired concentration?

M1V1=M2V2

b. 84.2 mL of a 4.60 M KMnO4 solution for a 1.42 M solution.

387.32

(4.60 M)(84.2ml) = ( 1.42M)(X ml)

1.42= 273 ml

8. What volume of the indicated solution is needed to produce the volume and concentration of a diluted solution

as indicated?M1V1=M2V2

a. 2.73 M NaOH solution to prepare 142 mL of a 0.540 M solution.

76.68

(2.73 M)(X ml) = ( 0.540M)(142 ml)

2.73= 28.1 ml

8. What volume of the indicated solution is needed to produce the volume and concentration of a diluted solution

as indicated?M1V1=M2V2

b. 0.0076 M SnF2 solution to prepare 25.0 mL of a 0.00027 M solution.

0.00675

(0.0076 M)(X ml) = ( 0.00027M)(25.0 ml)

0.0076= 0.888 ml

Ice Cream Lab

• 8 cups of whole milk• 8 cups of heavy whipping cream• 4 teaspoon vanilla• 4 cups of sugar

• Then the Dry Ice• Let’s do it- and eat

DO NOT CHEW ANY ICE!!!!Get a chunk-

spit it out