InClass01 105F96

Chemistry 105

InClass Problem Set #1Answer Sheet

29 Aug 1996 Name ______________________________

1. Reagent grade aqueous ammonia is 14.5 M. Directions call for using 5 mL of 6M NH3 to clean each of your three sintered glass crucibles. How will you make the 6 M NH3?

A. How much 6 M NH3 should you make up? Why?

3 crucibles x 5 mL each = 15 mL

B. How many moles of NH3 will the solution in "A" contain?

15 mL

Liter
6 mole

1000 mL
Liter
= 0.090 moles of NH3

C. How many mL of the 14.5 M NH3 solution will we need?

0.090 moles

Liter
1000 mL

14.5 moles
Liter
= 6.21 mL

D. What is the procedure for making the 6 M NH3?

Add 6.2 mL of 14.5 M NH3 to about 15 - 6.2 = 8.8 mL of H2O. [More accurate solutions require a volumetric flask.]

2. How many grams of AgNO3 are required to make 500 mL of 0.1M AgNO3?

A. How many moles of AgNO3 are in 500 mL of 0.1M AgNO3?

0.500 L

0.1 mole AgNO3

L
= 0.0500 moles of AgNO3

B. How many grams will these moles weigh? (The formula weight of AgNO3 is 169.837 grams/mole)

0.0500 moles AgNO3

169.837 grams

mole
= 8.49 grams

3. How much 0.1 M AgNO3 will be required to precipitate 0.2000 g of unknown chloride, if the sample is 60% chloride? (atomic wt. of Cl is 35.453)

A. How many grams of chloride are in the sample?

0.200 g unknown

0.60 g Cl-

g unknown
= 0.120 g Cl-

B. How many moles of chloride are in the sample?

0.12 g Cl-

mole Cl-

35.453 g Cl-
= 3.385 x 10-3 moles Cl-

C. How many moles of silver are required to precipitate the chloride? (Hint: Write the balanced chemical reaction between Ag+ and Cl-.)

Ag+ + Cl- = AgCl (s)

3.385 x 10-3 moles Cl-

1 mole Ag+

1 mole Cl-
= 3.385 x 10-3 moles Ag+

D. How many mL of 0.1M AgNO3 are required to precipitate the chloride?

3.385 x 10-3 moles Ag+

liter
1000 mL

0.1 moles Ag+
liter

= 33.9 mL

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