**Solutins****
to Exercises p 62-63**

1. A
technician needs 2.0 L of a 1.8 g/L solution of HNO_{3}. Sitting on the
counter is concentrated HNO_{3} (10 g/L ) .
How much of the 10 M solution should he carefully
dilute to 2.0 L ?

C_{1}V_{1} = C_{2}V_{2}

10V_{1} = 1.8(2)

V_{1} = 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?

C_{1}V_{1}
= C_{2}V_{2}

2V_{1} = 0.50(0.250)

V_{1} = 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?

C_{1}V_{1}
= C_{2}V_{2}

4V_{1} = 1(0.10)

V_{1} = 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.

C_{1}V_{1}
= C_{2}V_{2}

C_{1}(0.025) = 3.0(0.300)

C_{1 }= 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?

C_{1}V_{1}
= C_{2}V_{2}

3(0.050) = C_{2} (0.050
+0.250)

C_{2
}= 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?

C_{1}V_{1}
= C_{2}V_{2}

6.5(0.020) = 2.8 V_{2}

V_{2
}= 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?

C_{1}V_{1}
= C_{2}V_{2}

3.0(1.5) = C_{2} (1.5
+0.200)

C_{2
}= 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. C_{1}V_{1} = C_{2}V_{2}

1 V_{1}

V_{1}

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) =0.50V_{2}

V_{2}

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

14. C_{1}V_{1} = C_{2}V_{2}

45(0.060)= 30V_{2}

V_{2}

15. C_{1}V_{1} = C_{2}V_{2}

60(0.200) = C_{2} (0.200+0.400)

C_{2} = 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