Solutins to Exercises p 62-63

1.         A technician needs 2.0 L of a 1.8 g/L solution of HNO3. Sitting on the counter is concentrated HNO3 (10 g/L ) . How much of the 10 M solution should he         carefully dilute to 2.0 L ?

C1V1 = C2V2

10V1 = 1.8(2)

V1 = 0.36 L

2.         You want to prepare 250 mL of a 0.50 g/L solution from a 2.0 g/L solution.

How many mL should you pipet from the 2.0 g/L solution?

C1V1 = C2V2

2V1 = 0.50(0.250)

V1 = 0.0625 L = 62.5 mL

3.         a.         0.75 L of a 4 g/L solution are on the counter. How much of the solution should he dilute to 0.10 L to make a 1 g/L solution?

C1V1 = C2V2

4V1 = 1(0.10)

V1 = 0.025 L = 25 mL

b.                  Outline the lab procedure.

PTA(see notes)

4.         A 25.0 mL pipette is available. You want to end up with 300 mL of a 3.0 g/L solution. How concentrated should your original solution be if 25.0 mL will be used for dilution.

C1V1 = C2V2

C1(0.025) = 3.0(0.300)

C1 = 36g/L

5.         To 50 ml of a 3g/L solution, a student added 250 ml of water. What was the final

concentration of the solution?

C1V1 = C2V2

3(0.050) = C2 (0.050 +0.250)

C2 = 0.50 g/L

6.         How much water should be added to 20 .0 mL of a 6.5 g/L solution in order to create a 2.8 g/L solution?

C1V1 = C2V2

6.5(0.020) = 2.8 V2

V2 = 0.046 L

So he should add 0.046 – 0.020 = 0.026 L of water

7.         (430 only) What is the concentration of a solution created by adding 200 mL of water to 1.5 L of a 3.0 mole/L solution?

C1V1 = C2V2

3.0(1.5) = C2 (1.5 +0.200)

C2 = 2.6 g/L

8.         (430 only) In 100 mL of a solution, there are 3.0 g of NaCl.  Find the molarity.

3.0g/(58.5 g/mole) = 0.051 moles

C = n/V = 0.051 moles/0.100 L = 0.51 mole/L

9.         How many grams of KBr are  needed to prepare 2.5 L of a 0.25 g/L solution?

0.625 g

10.       How many grams of Ca(ClO)2 are needed to prepare 2.0 L of a 0.45 g/L

solution?

0.90 g

11.       Explain how you would actually prepare 3.0 L of a 0.2 g/L  Na Br solution in the lab.

1.                  Using a balance, weigh out 0.6 g of NaBr crystals.

2.                  Dissolve in less than 3.0 L in a beaker.

3.                  Transfer to a 3.0 L volumetric flask. Rinse beaker into flask.

4.                  Dilute to flask’s white mark with water.Mix.

12. C1V1 = C2V2

1 V1 = 0.01(0.5)

V1 =0.050 L

Notice that the 4.0 L is useless. We're trying to find what volume from the 4.0 L we need to take out for dilution.

13.

2(0.30) =0.50V2

V2 =1.2 L

so we need to add 1.2L -0.30 L = 0.90 L of water have to be added.

14. C1V1 = C2V2

45(0.060)= 30V2

V2 = 0.090 L = 90 ml

15. C1V1 = C2V2

60(0.200) = C2 (0.200+0.400)

C2 = 20 g/L

16. n = CV = 0.10(0.03525) = 0.003525 molesCa(OH)2

0.003525 moles Ca(OH)2( 6 mol water/3molCa(OH)2) = 0.00705 mol of water

0.00705 mol of water(18g/mol) =0.1269 g (ml/1.00g) = 0.1269 ml