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
V1
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
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
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