**Problem #1:** A concentrated ammonia solution is 27.0 % (m/m) NH_{3}. The density of the solution is 0.900 g/mL. What is the molarity of the solution?

**Solution:**

1) 27.0 % (m/m) means:

100.0 g of solution contains 27.0 g of NH_{3}

2) Calculate volume of 100.0 g of solution:

100.0 g ÷ 0.900 g/mL = 111.11 mL of solution

3) Determine moles of NH_{3}:

27.0 g ÷ 17.031 g/mol = 1.585344 mol (I kept a few guard digits.)

4) Determine the molarity:

1.585344 mol ÷ 0.11111 L = 14.3 M (to three sf)

Yes, this molarity is possible with NH_{3}. It's very soluble!

**Problem #2:** Sea water contains 3.90 x 10¯^{6} ppm of dissolved gold. What volume of this sea water would contain 1.00 g of gold?

**Solution:**

1) What does "ppm" mean?

The expression "1 ppm" means a given property exists at a relative proportion of one part per million parts examined, as would occur if a water-borne pollutant was present at a concentration of one-millionth of a gram per gram of sample solution.

2) Therefore, 3.90 x 10¯^{6} ppm means this:

3.90 x 10¯^{6}g of Au per 1.00 gram of seawater

3) We use a ratio and proportion:

(3.90 x 10¯^{6}g of Au ÷ 1.00 gram of seawater) = (1.00 g of Au ÷ x)

4) Cross multiply and divide yields our answer:

2.56 x 10^{5}g of sea water contains 1.00 g of gold

**Problem #3:** A solution used to chlorinate a home swimming pool contains 7% chlorine by mass. An ideal chlorine level for the pool is one part per million chlorine (think of 1 ppm as being 1 g chlorine per million grams of water). If you assume densities of 1.10 g/mL for the chlorine solution and 1.00 g/mL for the swimming pool water, what volume of the chlorine solution in liters, is required to produce a chlorine level of 1.00 ppm in an 18,000 gallon swimming pool?

**Solution:**

1) Convert 18,000 gallons to liters:

The conversion factor we will use is 1 gallon = 3.7854 L18,000 gal x 3.7854 L/gal = 6.81372 x 10

^{4}L

2) Determine how many grams of pool water this is:

6.81372 x 10^{7}mL x 1.00 g/mL = 6.81372 x 10^{7}gNote change from L to mL.

3) At 1 ppm, how much chlorine is required? Use a ratio and proportion:

(1 g chlorine ÷ 10^{6}g pool water) = (x ÷ 6.81372 x 10^{7}g of pool water)x = 68.1372 g chlorine required

4) What amount of 7% (by mass) chlorine solution is required to deliver 68.1372 g of chlorine?

(68.1372 g ÷ 0.07) = (x/1)x = 973.3886 g of chlorine solution required

5) What volume (in liters) is this?

973.3886 g ÷ 1.10 g/mL = 884.8987 mLTo three sig figs (which seems reasonable to the ChemTeam), the answer is 0.885 L.

**Problem #4:** How many ions are in 20.0 micro liters of 0.00100 micrograms/mL dilution of NaCl?

**Solution:**

1) Convert 20.0 μL to mL:

20.0 μL x (1 L / 10^{6}μL) = 20.0 x 10¯^{6}L20.0 x 10¯

^{6}L x (1000 mL/L) = 0.0200 mL

2) Determine grams of NaCl in 0.0200 mL

(0.00100 μg/mL) x 0.0200 mL = 2.00 x 10¯^{5}μg2.00 x 10¯

^{5}μg x (1 g / 10^{6}μg) = 2.00 x 10¯^{11}g

3) Determine moles of NaCl:

2.00 x 10¯^{11}g / 58.443 g/mol = 3.4221378 x 10¯^{13}mol

4) Determine formula units of NaCl

(3.4221378 x 10¯^{13}mol) (6.022 x 10^{23}mol¯^{1}) = 2.0608 x 10^{11}

5) Determine ions of NaCl:

2.0608 x 10^{11}times 2 = 4.12 x 10^{11}(to 3 sf)

**Problem #5:** 25.0 cm^{3} of a solution of 0.200 M potassium hydroxide reacts with 30.0 cm^{3} of a solution of nitric acid. What is the concentration of the acid in g dm^{-3}?

**Solution:**

1) Determine molarity of nitric acid sample:

M_{1}V_{1}= M_{2}V_{2}(0.200 mol/L) (25.0 cm

^{3}) = (x) (30.0 cm^{3})x = 0.167 M

2) Determine moles, then grams of HNO_{3} in 30.0 cm^{3}:

Molarity = moles / volume0.167 M = x / 0.03000 L

x = 0.00500 mol

grams = (0.00500 mol) (63.0119 g/mol) = 0.315 g

3) Determine g dm^{-3}

0.315 g / 0.0300 dm^{3}= 10.5 g dm^{-3}Remember that one liter equals one cubic decimeter.