Principles of Organic Synthesis, Third Edition (Norman, Richard O.C.; Coxon, James M.)

Shelton Bank

State University of New York at Albany, Albany, NY 12222

J. Chem. Educ., 1995, 72 (7), p A152

DOI: 10.1021/ed072pA152.1

Publication Date: July 1995

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Organic Chemistry, 2 edition by Jonathan Clayden,

Organic Chemistry, 2 edition by Jonathan Clayden, Nick Greeves and Stuart Warren

Topics reaction, chapter, reactions, bond, acid, carbonyl, group, compounds, carbon, bonds, carbonyl group, double bond, leaving group, conjugate addition, lone pair, carbon atom, nmr spectrum, starting material, interactive mechanism, carbonyl compounds

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By SAJJAD HUSSAIN MIRANI
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Converting between "ppm" and molarity

Converting between "ppm" and molarity

The definition of parts per million:

1 g solute per 1,000,000 g solution
Now, divide both values by 1000 to get a new definition for ppm:

ppm = 0.001 g per 1,000 g solution
or:

ppm = 1 mg solute per 1 kg solution
Then, for an aqueous solution:

ppm = 1 mg solute per liter of solution
This last one works because the solution concentration is so low that we can assume the solution density to be 1.00 g/mL.

Also, it's this last modification of ppm (the mg/L one) that allows us to go to molarity (which has units of mol/L).

The best way to explain this is by doing some examples:

Problem #1: Convert 78 ppm of Ca2+ ions to mol/L

Solution:

1) By the last definition of ppm just above:

78 ppm = 78 mg Ca2+ / L of solution = 0.078 g/L
2) Divide by the atomic weight for calcium ion:

0.078 g/L divided by 40.08 g/mol = 00019 mol/L
Problem #2: Calculate the molarity of a dye concentration given the molar mass is of the dye 327 g/mol and a dye concentration of 2 ppm.

Solution:

1) Convert ppm to a gram-based concentration:

2 ppm = 2 mg dye / L of solution
2a) Using 0.002 g/L, caculate the molarity:

0.002 g/L divided by 327 g/mol = 6.1 x 10-6 M
2b) Using 2 mg/L, calculate the molarity

2 mg/L divided by 327,000 mg/mol = 6.1 x 10-6 M
You might want to go back to problem #1 and try out 78 mg/L with the atomic weight of calcium ion expressed as mg/mol instead of g/mol.

Problem #3: A solution is labeled 2.89 ppm and is made with a solute that has molar mass equal to 522 g/mol. What is the molarity of the solution?

Solution (straight from Yahoo Answers):

1) Unless specified otherwise, ppm usually refers to ppm by weight:

2.89 ppm = 2.89 g per 1,000,000 g (or any other weight unit, I have just chosen g for convenience)
2) Now you need to know the density of the solvent to convert the volume to mass. If the solvent is water, we can assume the density at standard temperature and pressure is 1.0 g/mL. Therefore 1 litre (L) of water is 1,000 g. Therefore:

2.89 ppm = 2.89 g per 1,000 L = 0.00289 g per 1 L
(If you work out the mass per litre it makes working out the next steps a little easier, because the final units for molarity will be mol/L)

3) Now using the molar mass to work out the molarity (moles per litre):

Moles in one litre = mass/molar mass = 0.00289/522 = 5.5 x 10-6 mol
Therefore the molarity is 5.5 x 10-6 M

Problem #4: What is the ppm concentration of calcium ion in 0.010 M CaCO3?

Solution:

1) Convert mol/L to to g/1000 g of solution:

0.010 mol/L times 40.08 g/mol = 0.4008 g/L
0.4008 g/L = 0.4008 g / 1000 g of solution

Note that the solution density is assumed to be 1.00 g/mL. This is the common practice in ppm calculations.

2) Convert to ppm by multiplying numerator and denominator by 1000:

0.4008 g / 1000 g of solution times 1000/1000 = 400 g / 1,000,000 g of solution
The answer is 400 ppm

Notice how the weight for calcium ion only is used. The weight of the carbonate, based on the wording of the question, is immaterial to the solving of the problem.

This focus on an ion (and not the entire formula of the substance) in ppm concentrations is common. Be aware!

Problem #5: Make a 20.0 ppm solution of HCl from a stock solution that is 0.500-molar.

Solution:

1) Let us assume we will make 1.00 L of solution. Since ppm = 1 mg solute per liter of solution:

we need 20.0 mg solute per liter of solution
2) Calculate the volume of 0.500 M solution that contains 20.0 mg HCl:

MV = g/molar mass
(0.500 mol/L) (x) = 0.020 g / 36.46 g/mol

x = 0.001097 L

We need 1.10 mL of the stock solution diluted to 1.00 L to make a 20.0 ppm solution of HCl.

3) A slightly different form of the above question was answered thusly on Yahoo Answers:

The unit "ppm" is equivalent to "mg/L."
Concentrated HCl is 12.3 moles/L or, since the molar mass of HCl is 36.5 g/L, 449 g HCl / L.

449 g/L = 449,000 mg/L. If you want 20 ppm, you need to make a 20/449,000 volume dilution, or 1/22,500.

The best way to do that is a serial dilution: first make a 1/100 dilution (1 mL conc. HCl / 100 mL) and then dilute THAT solution by 1/225. The result: 1/100 x 1/225 = 1/22,500.

You can cut the dilution process to one step if a lower concentration of HCl is available (e.g. 0.1 M).

Problem #6: In a geographical area at sea level measuring 56 km by 24 km by 1 km, the ozone concentration is 0.145 ppm. How many moles of ozone must be removed from the atmosphere to reach a concentration of 0.080 ppm, assuming a constant pressure of 1 atm and a constant temperature of 30 degrees Celsius.

Solution:

1) Calculate the liters (use cubic decimeters) in our volume:

(56 x 105 dm) (24 x 105 dm) (1 x 105 dm) = 1.344 x 1018 dm3
2) Calculate the ppm that must be removed:

0.145 - 0.080 = 0.065
3) Determine milligrams (then grams, then moles) of ozone to be removed from our volume:

1 ppm = 1 mg/L
Therefore:

0.065 mg/L times 1.344 x 1018 L = 8.736 x 1016 mg

8.736 x 1016 mg = 8.736 x 1013 g

8.736 x 1013 g divided by 47.997 g/mol = 1.75 x 1012 mol

Problem #7: 3100 mL of a river water sample took 9.30 mL of 0.01005 M Ag+ to titrate. Calculate the concentration of Cl¯ in ppm for the river water.

Solution:

1) Calculate moles of Ag+:

(0.01005 mol/L) (0.00930 L) = 9.3465 x 10-5 mol of Ag+
2) Since Ag+ and Cl¯ react in a 1:1 molar ratio:

9.3465 x 10-5 mol of Cl¯
3) How many mg of Cl¯ is this?

9.3465 x 10-5 mol x 35.453 g/mol = 3.31 x 10-3 g
3.31 x 10-3 g x (1000 mg/g) = 3.31 mg

4) ppm = mg/L

3.31 mg / 3.100 L = 1.07 ppm (to three sf)
Problem #8: A solution contains 6.0 x 10-6 mol Na2SO4 in 250.0 mL. What is the concentration of sodium ion in ppm? Of sulfate ion?

Solution:

1) Convert to molarity:

6.0 x 10-6 mol / 0.250 L = 2.40 x 10-5 mol/L
2) Convert mol/L to grams of sodium ion per 1000 g of solution:

2.40 x 10-5 mol/L times 23.00 g/mol x 2 = 0.001104 g/L
0.001104 g/L = 0.001104 g/1000 g of solution

The two is used because there are two sodium ions per formula unit of sodium sulfate.

For sulfate, the two would not be used and, instead of 23.00 g/mol, use the "formula weight" of the sulfate ion.

3) Convert to ppm by multiplying numerator and denominator by 1000:

0.001104 g/1000 g of solution times 1000/1000 = 1.104 g / 1,000,000 g of solution
The answer (for sodium ion) is 1.1 ppm (to two sig figs).

The solution for sulfate is left to the reader.

Problem #8: The maximum concentration of O2 in seawater is 2.2 x 10-4 M at 25.0 °C. What is this concentration in ppm?

Solution:

1) Convert mol/L to grams of sodium ion per 1000 g of solution:

2.2 x 10-4 mol/L times 32.0 g/mol = 0.00704 g/L
0.00704 g/L = 0.00704 g/1000 g of solution

2) Convert to ppm by multiplying numerator and denominator by 1000:

0.00704 g/1000 g of solution times 1000/1000 = 7.04 g / 1,000,000 g of solution
The answer is 7.0 ppm (to two sig figs).

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What does "ppm" mean?

What does "ppm" mean?
                                                        Go To Converting between "ppm" and molarity
                                                                       
The expression "1 ppm" means a given solute exists at a concentration of one part per million parts of the solution. These are two common ways to think about what the concentration "1 ppm" means:

a) it is one-millionth of a gram per gram of sample solution.
b) it is one gram of solute per million grams of sample solution.

Notice that the more general word 'part' is used above, but 'gram' is used in (a) and (b) just above. This is because 'gram' is used almost exclusively when parts per million is used.

The best way to explain this is by doing some examples:

Problem #1: 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) 3.90 x 10¯6 ppm means this:

3.90 x 10¯6 g of Au per 1.00 gram of seawater
2) 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)
3) Cross multiply and divide yields our answer:

2.56 x 105 g of sea water contains 1.00 g of gold
Problem #2: Pollutants in air and water are frequently measured in parts per million (ppm) or parts per billion (ppb). One part per million would mean that there is one gram of the pollutant in every one million grams of air. At ordinary temperature and pressure, air has a density of 0.00012 gram per cubic centimeter. What volume of air would contain one gram of sulfur dioxide, a pollutant that causes acid rain, if the sulfur dioxide concentration is 2 ppm.

Solution:

1) How many grams of air contain one gram of SO2?

2 ppm means 2 grams SO2 per million grams of air
therefore,

one gram of SO2 will be found in 500,000 g of air

2) What volume of air weighs 500,000 g? We use the density (notice I set it up in a ratio and proportion style:

0.00012 g over 1 cm3 = 500,000 g over x
x = 4.17 x 109 cm3

Problem #3: A sample of oil (density = 0.89 g/mL) was found to have dioxin contamination of 2 ppm. How many mL of the oil would contain 0.01 gram of dioxin?

Solution:

1) Determine grams of oil holding 0.01 g of dioxin. Use a ratio and proportion:

(2 g dioxin over 1,000,000 g solution) = (0.01 g over x)
x = 5000 g

2) Determine the volume of oil:

0.89 g/mL = 5000 g divided by x
x = 5618 mL

Problem #4: Fish need about 5 ppm O2 dissolved in water to survive. Will water with 7 mg O2 per liter sustain fish?

Solution:

We need to convert 7 mg/L into ppm:

(0.007 g / 1000 g) = (x / 1,000,00 g)
x = 7 ppm

The answer is yes.

Notice the conversion of one liter of water (with 7 mg dissolved oxygen) into 1000 g. I used the density of 1.00 g/mL for the conversion. This is acceptable because solutions in the ppm range are so dilute that the solute (in this case, the O2, has no effect on the solution density.

Look again at the mg/L value. Another way to explain getting from mg/L to ppm is to multiply both numerator and denominator by 1000:

(0.007 g / 1000 g) x (1000 / 1000) = 7 g / 1,000,000 g = 7 ppm
You may very well face problems where you get the data in mg/L format and are then asked something about ppm. Remember!

Problem #5: Show why a mass of 0.145 g of KMnO4 in a 500 mL volumetric flask corresponds to 100 ppm in Mn.

Solution:

1) Using a gravimetric factor, determine grams of Mn in 0.145 g of KMnO4:

0.145 g times (54.938 g/mol over 158.032 g/mol) = 0.0504076 g
2) Calculate ppm of Mn. Use a ratio and proportion:

(0.050 g over 500 g) = (x over 1,000,000)
x = 100 ppm

Problem #6: Symptoms of lead poisoning become apparent after a person has accumulated more than 20 mg in the body. Express this amount as parts per million for an 80 kg person.

Solution:

1) Convert mg and kg to grams:

20 mg = 20 x 10-3 g
80 kg = 80 x 103 g
2) Determine grams Pb per g of bodyweight:

20 x 10-3 g / 80 x 103 g = 0.25 x 10-6 per g of bodyweight
3) Refer to first description of ppm given at beginning of file:

one ppm "is one-millionth of a gram per gram of sample solution."
In this case, the "solution" is the bodyweight.

answer = 0.25 ppm

Problem #7: If you eat a 6 oz. can (180g) of tuna with 0.20 ppm Hg, how much mercury do you ingest?

Solution:

1) The meaning of 0.20 ppm Hg:

Remember, one ppm "is one-millionth of a gram per gram of sample solution."
Therefore the tuna contains 0.20 x 10-6 g of Hg per gram of tuna.

2) Calculate Hg in 180 g of tuna:

0.20 x 10-6 g of Hg per g of tuna times 180 g of tuna = 3.6 x 10-5 g of Hg
Problem #8: It is estimated that 3 x 105 tons of sulfur dioxide enters the atmosphere daily owing to the burning of coal and petroleum products. Assuming an even distribution of the sulfur dioxide throughout the earth's atmosphere (which is not the case), calculate in parts per million by weight the concentration of SO2 added daily to the atmosphere. The weight of the atmosphere is 4.5 x 1015 tons.

Solution:

1) Set up a ratio and proportion:

You are adding 3 x 105 parts per 4.5 x 1015 parts.
How many parts per 1 x 106 is this?

2) Solve it:

3 x 105 parts is to 4.5 x 1015 parts as x is to 1 x 106 parts
x = 0.000067 ppm

Problem #9: 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. 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 L
18,000 gal x 3.7854 L/gal = 6.81372 x 104 L

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

6.81372 x 107 mL x 1.00 g/mL = 6.81372 x 107 g
Note change from L to mL.

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

(1 g chlorine ÷ 106 g pool water) = (x ÷ 6.81372 x 107 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 mL
To three sig figs (which seems reasonable to the ChemTeam), the answer is 0.885 L.

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Molarity Problems

Molarity Problems         
                                                                         Molarity Difination
The equations I will use are:

M = moles of solute / liters of solution
and

MV = grams / molar mass <--- The volume here MUST be in liters.
Typically, the solution is for the molarity (M). However, sometimes it is not, so be aware of that. A teacher might teach problems where the molarity is calculated but ask for the volume on a test question.

Note: Make sure you pay close attention to multiply and divide. For example, look at answer #8. Note that the 58.443 is in the denominator on the right side and you generate the final answer by doing 0.200 times 0.100 times 58.443.

Problem #1: Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water?

Solution:

MV = grams / molar mass
(x) (1.00 L) = 28.0 g / 58.443 g mol¯1

x = 0.4790993 M

to three significant figures, 0.479 M

Problem #2: What is the molarity of 245.0 g of H2SO4 dissolved in 1.000 L of solution?

Solution:

MV = grams / molar mass
(x) (1.000 L) = 245.0 g / 98.0768 g mol¯1

x = 2.49804235 M

to four sig figs, 2.498 M

If the volume had been specified as 1.00 L (as it often is in problems like this), the answer would have been 2.50 M, NOT 2.5 M. You want three sig figs in the answer and 2.5 is only two SF.

Problem #3: What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?

Solution:

MV = grams / molar mass
(x) (0.4000 L) = 5.30 g / 105.988 g mol¯1

0.12501415 M

x = 0.125 M (to three sig figs)

Problem #4: What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?

Solution:

MV = grams / molar mass
(x) (0.7500 L) = 5.00 g / 39.9969 g mol¯1

(x) (0.7500 L) = 0.1250097 mol <--- threw in an extra step

x = 0.1666796 M

x = 0.167 M (to three SF)

Problem #5: How many moles of Na2CO3 are there in 10.0 L of 2.00 M solution?

Solution:

M = moles of solute / liters of solution
2.00 M = x / 10.0 L

x = 20.0 mol

Suppose the molarity was listed as 2.0 M (two sig figs). How to display the answer? Like this:

20. mol

Problem #6: How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution?

Solution:

M = moles of solute / liters of solution
2.0 M = x / 0.0100 L <--- note the conversion of mL to L

x = 0.020 mol

Problem #7: How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?

Solution:

0.200 M = x / 0.1000 L
x = 0.0200 mol

Problem #8: What weight (in grams) of NaCl would be contained in problem #7?

Solution:

(0.200 mol L¯1) (0.100 L) = x / 58.443 g mol¯1 <--- this is the full set up
x = 1.17 g (to three SF)

You could have done this as well:

58.443 g/mol times 0.0200 mol <--- this is based on knowing the answer from problem #7

Problem #9: What weight (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?

Solution:

(2.00 mol L¯1) (0.7500 L) = x / 98.0768 g mol¯1
x = (2.00 mol L¯1) (0.7500 L) (98.0768 g mol¯1)

x = 147.1152 g

to three sig figs, 147 g

Problem #10: What volume (in mL) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?

Solution:

(18.0 mol L¯1) (x) = 2.45 g / 98.0768 g mol¯1
(18.0 mol L¯1) (x) = 0.0249804235 mol

x = 0.0013878 L

The above is the answer in liters. Multiplying the answer by 1000 provides the required mL value:

0.0013878 L times (1000 mL / L) = 1.39 mL (given to three sig figs)

Problem #11: What volume (in mL) of 12.0 M HCl is needed to contain 3.00 moles of HCl?
Solution:

12.0 M = 3.00 mol / xx = 0.250 L
This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value:
0.250 L x (1000 mL / L) = 250. mL (note use of explicit decimal point to create three sig figs)

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Problem #12: How many grams of Ca(OH)₂ are needed to make 100.0 mL of 0.250 M solution?
Solution:

(0.250 mol L¯¹) (0.100 L) = x / 74.0918 g mol¯¹x = (0.250 mol L¯¹) (0.100 L) (74.0918 g mol¯¹)
x = 1.85 g (to three sig figs)

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Problem #13: What is the molarity of a solution made by dissolving 20.0 g of H₃PO₄ in 50.0 mL of solution?
Solution:

(x) (0.0500 L) = 20.0 g / 97.9937 g mol¯¹(x) (0.0500 L) = 0.204094753 mol
x = 4.08 M

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Problem #14: What weight (in grams) of KCl is there in 2.50 liters of 0.500 M KCl solution?
Solution:

(0.500 mol L¯¹) (2.50 L) = x / 74.551 g mol¯¹x = 93.2 g

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Problem #15: What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?
Solution:

(x) (0.2500 L) = 12.0 g / 39.9969 g mol¯¹ <--- note use of L in set up, not mL x = 1.20 M

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Problem #16: Determine the molarity of these solutions:

a) 4.67 moles of Li₂SO₃ dissolved to make 2.04 liters of solution.
b) 0.629 moles of Al₂O₃ to make 1.500 liters of solution.
c) 4.783 grams of Na₂CO₃ to make 10.00 liters of solution.
d) 0.897 grams of (NH₄)₂CO₃ to make 250 mL of solution.
e) 0.0348 grams of PbCl₂ to form 45.0 mL of solution.

Solution set-ups:

a) x = 4.67 mol / 2.04 L
b) x = 0.629 mol / 1.500 L
c) (x) (10.00 L) = 4.783 g / 106.0 g mol¯¹
d) (x) (0.250 L) = 0.897 g / 96.09 g mol¯¹
e) (x) (0.0450 L) = 0.0348 g / 278.1 g mol¯¹

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Problem #17: Determine the number of moles of solute to prepare these solutions:

a) 2.35 liters of a 2.00 M Cu(NO₃)₂ solution.
b) 16.00 mL of a 0.415-molar Pb(NO₃)₂ solution.
c) 3.00 L of a 0.500 M MgCO₃ solution.
d) 6.20 L of a 3.76-molar Na₂O solution.

Solution set-ups:

a) x = (2.00 mol L¯¹) (2.35 L)
b) x = (0.415 mol L¯¹) (0.01600 L)
c) x = (0.500 mol L¯¹) (3.00 L)
d) x = (3.76 mol L¯¹) (6.20 L)

Comment: the technique used is this:

MV = moles of solute

This particular variation of the molarity equation occurs quite a bit in certain parts of the acid base unit.

---

Problem #18: Determine the grams of solute to prepare these solutions:

a) 0.289 liters of a 0.00300 M Cu(NO₃)₂ solution.
b) 16.00 milliliters of a 5.90-molar Pb(NO₃)₂ solution.
c) 508 mL of a 2.75-molar NaF solution.
d) 6.20 L of a 3.76-molar Na₂O solution.
e) 0.500 L of a 1.00 M KCl solution.
f) 4.35 L of a 3.50 M CaCl₂ solution.

Solution set-ups:

a) (0.00300 mol L¯¹) (0.289 L) = x / 187.56 g mol¯¹
b) (5.90 mol L¯¹) (0.01600 L) = x / 331.2 g mol¯¹
c) (2.75 mol L¯¹) (0.508 L) = x / 41.99 g mol¯¹
d) (3.76 mol L¯¹) (6.20 L) = x / 61.98 g mol¯¹
e) (1.00 mol L¯¹) (0.500 L) = x / 74.55 g mol¯¹
f) (3.50 mol L¯¹) (4.35 L) = x / 110.99 g mol¯¹

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Problem #19: Determine the final volume of these solutions:

a) 4.67 moles of Li₂SO₃ dissolved to make a 3.89 M solution.
b) 4.907 moles of Al₂O₃ to make a 0.500 M solution.
c) 0.783 grams of Na₂CO₃ to make a 0.348 M solution.
d) 8.97 grams of (NH₄)₂CO₃ to make a 0.250-molar solution.
e) 48.00 grams of PbCl₂ to form a 5.0-molar solution.

Solution set-ups:

a) x = 4.67 mol / 3.89 mol L¯¹
b) x = 4.907 mol / 0.500 mol L¯¹
c) (0.348 mol L¯¹) (x) = 0.783 g / 105.99 g mol¯¹
d) (0.250 mol L¯¹) (x) = 8.97 g / 96.01 g mol¯¹
e) (5.00 mol L¯¹) (x) = 48.0 g / 278.1 g mol¯¹

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Problem #20: A student placed 11.0 g of glucose (C₆H₁₂O₆) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500L. How many grams of glucose are in 100. mL of the final solution?
Solution path #1:
1) Calculate molarity of first solution (produced by dissolving 11.0 g of glucose):

MV = grams / molar mass(x) (0.100 L) = 11.0 g / 180.155 g/mol
x = 0.610585 mol/L (I'll carry a few guard digits.)

2) Calculate molarity of second solution (produced by diluting the first solution):

M₁V₁ = M₂V₂(0.0200 L) (0.610585 mol/L) = (0.500 L) (x)
x = 0.0244234 mol/L

3) Determine grams of glucose in 100. mL of second solution:

MV = grams / molar mass(0.0244234 mol/L) (0.100 L) = x / 180.155 g/mol
x = 0.44 g

Solution path #2:
1) Calculate how much glucose you have in 20.0 mL of the first solution.

11.0 g is to 100. mL as x is to 20.0 mLCross-multiply and divide
100x = 11.0 times 20.0
x = 2.2 g

2) When you dilute the 20.0 mL sample to 500.0 mL, you have 2.2 g glucose in the solution.

2.2 g is to 500. mL as x is to 100. mLCross-multiply and divide
500x = 2.2 times 100
x = 0.44 g
In 100 mL of the final solution are 0.44 g glucose.

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Problem #21: Commercial bleach solution contains 5.25% (by mass) of NaClO in water. It has a density of 1.08 g/mL. Caculate the molarity of this solution. (Hints: assume you have 1.00 L of solution; molar mass of NaClO 74.4 g/mol)
Solution:
1) Determine mass of 1.00 L of solution:

(1.08 g/mL) (1000 mL) = 1080 g

2) Determine mass of NaClO in 1080 g of solution:

(1080 g) (0.0525) = 56.7 g

3) Determine moles of NaClO:

56.7 g / 74.4 g/mol = 0.762 mol

4) Determine molarity of solution:

0.762 mol / 1.00 L = 0.762 M

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Problem #22: What is the molality (and molarity) of a 20.0% by mass hydrochloric acid solution? The density of the solution is 1.0980 g/mL.
Solution: 1) Determine moles of HCl in 100.0 g of 20.0% solution.

20.0 % by mass means 20.0 g of HCl in 100.0 g of solution.20.0 g / 36.4609 g/mol = 0.548 mol

2) Determine molality:

0.548 mol / 0.100 kg = 5.48 m

3) Determine volume of 100.0 g of solution.

100.0 g / 1.0980 g/mL = 91.07468 mL

4) Determine molarity:

0.548 mol / 0.09107468 L = 6.02 m

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Problem #23: 25.0 mL of 0.250 M KI, 25.0 mL of 0.100 K₂SO₄, and 15.0 mL of 0.100 M MgCl₂ were mixed together in a beaker. What are the molar concentrations of I¯, Cl¯ and K⁺ in the beaker?
Solution:
1) Calculate the total volume of the mixed solutions:

25.0 mL + 25.0 mL + 15.0 mL = 65.0 mL

2) Concentration of iodide ion:

M₁V₁ = M₂V₂(0.250 mol/L) (25.0 mL) = (x) (65.0 mL)
x = 0.09615 M
to three sig figs, 0.0962 M

3) Concentration of the chloride ion:

moles Cl¯ ---> (0.100 mol/L) (0.0150 L) (2 Cl¯ / 1 MgCl₂) = 0.00300 mol0.00300 mol / 0.065 L = 0.04615 M
to three sig figs, 0.0462 M

4) Concentration of the potassium ion:

moles K⁺ from KI ---> (0.250 mol/L) (0.0250 L) = 0.00625 mol
moles K⁺ from K₂SO₄ ---> (0.100 mol/L) (0.0250 L) (2 K⁺ / 1 K₂SO₄) = 0.00500 mol0.00625 mol + 0.00500 mol = 0.01125 mol
0.01125 mol / 0.0650 L = 0.173 M

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Problem #24: Calculate the total concentration of all the ions in each of the following solutions:

a. 3.25 M NaCl
b. 1.75 M Ca(BrO₃)₂
c. 12.1 g of (NH₄)₂SO₃ in 615 mL in solution.

Solution:
1) the sodium chloride solution:

for every one NaCl that dissolves, two ions are produced (one Na⁺ and one Cl¯).the total concentration of all ions is this:
(3.25 mol/L) times (2 total ions / 1 NaCl formula unit) = 6.50 M

2) the calcium bromate solution:

three total ions are produced for every one Ca(BrO₃)₂ that dissolves (one Ca²⁺ and two BrO₃¯the total concentration of all ions is this:
(1.75 mol/L) times (3 total ions / 1 Ca(BrO₃)₂ formula unit) = 5.25 M

3) the ammonium sulfite solution:

calculate the concentration of (NH₄)₂SO₃:(x) (0.615 L) = 12.1 g / 116.1392 g/mol
x = 0.169407 M <--- I'll carry some guard digits
calculate the concentration of all ions:
(NH₄)₂SO₃ produces three ions for every one formula unit that dissolves.
0.169407 M times 3 = 0.508221 M
to three sig figs, 0.508 M

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Problem #25: A solution of calcium bromide contains 20.0 g dm^-3. What is the molarity of the solution with respect to calcium bromide and bromine ions.
Solution:

MV = mass / molar mass(x) (1.00 L) = 20.0 g / 199.886 g/mol
x = 0.100 M
When CaBr₂ ionizes, two bromide ions are released for every one CaBr2 that dissolves. That leads to this:
[Br^-] = 0.200 M

Problem #26: What is the concentration of each type of ion in solution after 23.69 mL of 3.611 M NaOH is added to 29.10 mL of 0.8921 M H₂SO₄? Assume that the final volume is the sum of the original volumes.
Solution:
The answer requires you to know how NaOH and H₂SO₄ react. Here is the chemical equation:

H₂SO₄ + 2NaOH ---> Na₂SO₄ + 2H₂O

A key point is that two NaOH formula units are required for every one H₂SO₄
1) Calculate moles of NaOH and H₂SO₄:

moles NaOH ---> (3.611 mol/L) (0.02369 L) = 0.08554459 mol
moles H₂SO₄ ---> (0.8921 mol/L) (0.02910 L) = 0.02596011 mol

2) Determine how much NaOH remains after reacting with the H₂SO₄:

0.02596011 mol x 2 = 0.05192022 mol <--- moles of NaOH that react0.08554459 mol - 0.05192022 mol = 0.03362437 mol <--- moles of NaOH that remain

3) The above was required to determine the hydroxide ion concentration:

0.03362437 mol / 0.05279 L = 0.6369 M0.05279 L is the sum of the two solution volumes.

4) Determine the sodium ion concentration:

0.08554459 mol / 0.05279 L = 1.620 MThe sole source of sodium ion is from the NaOH.

5) Determine the sulfate ion concentration:

0.02596011 mol / 0.05279 L = 0.4918 MThe sole source of sulfate ion is from the H₂SO₄.

6) Determine the hydrogen ion concentration:

[H⁺] [OH¯] = 1.000 x 10^-14(x) (0.636945823) = 1.000 x 10^-14
x = 1.570 x 10^-14 M

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Problem #27: Given 3.50 mL of sulfuric acid (98.0% w/w) calculate the number of mmols in the solution (density: 1.840 g/mL).
Solution:

3.50 mL times 1.840 g/mL = 6.44 g <--- mass of the 3.50 mL6.44 g times 0.980 = 6.3112 g <--- mass of H₂SO₄ in the solution
6.3112 g / 98.0768 g/mol = 0.06434957 mol
0.06434957 mol times (1000 mmol / 1 mol) = 64.3 mmol (to three sig figs)

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Problem #28: Given 8.00 g of HBr calculate the volume (mL) of a 48.0% (w/w) solution. (MW HBr: 80.9119 g/mol, density: 1.49 g/mL). Then, calculate the molarity.
Solution:

8.00 g divided by 0.48 = 16.6667 g <--- total mass of the solution in which the HBr is 48% by mass16.6667 g divided by 1.49 g/mL = 11.18568 mL
to three sig figs, the volume of the solution is 11.2 mL
For the molarity, determine the moles of HBr:
8.00 g / 80.9119 g/mol = 0.098873 mol
0.098873 mol / 0.01118568 L = 8.84 M

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Problem #29: A solution is made by dissolving 0.100 mol of NaCl in 4.90 mol of water. What is the mass % of NaCl?
Solution:
1) Convert moles to masses:

NaCl ---> 0.100 mol times 58.443 g/mol = 5.8443 gH₂O ---> 4.90 mol times 18.015 g/mol = 88.2735 g

2) Calculate mass percent of NaCl:

[5.8443 g / (5.8443 g + 88.2735 g)] * 100 = 6.21% (to three sig figs)

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Problem #30: 2.00 L of HCl gas (measured at STP) is dissolved in water to give a total volume of 250. cm³ of solution. What is the molarity of this solution?
Solution using molar volume:

2.00 L divided by 22.414 L/mol = 0.0892299 mol of HCl0.0892299 mol / 0.250 L = 0.357 M (to three sig figs)

Solution using Ideal Gas Law:

PV = nRT(1.00 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (273.15 K)
n = 0.0892272 mol
0.0892272 mol / 0.250 L = 0.357 M (to three sig figs)

If the volume of HCl gas was not at STP, you must use PV = nRT to calculate the moles. You cannot use molar volume since it is only true at STP.

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Bonus Problem: How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L of a solution that has a pH of 2.01?
Solution:
1) Get moles of hydrogen ion needed for the 18.0 L:

[H⁺] = 10^-pH = 10^-2.01 = 0.0097724 M0.0097724 mol/L) (18.0 L) = 0.1759032 mol of HCl required
Remember, HCl is a strong acid, dissociating 100% in solution

2) Determine molarity of 36.0% HCl:

Assume 100. g of solution present.36.0 g of that is HCl
100. g / 1.18 g/mL = 84.745763 mL
Use MV = mass / molar mass
(x) (0.084745763 L) = 36.0 g / 36.4609 g/mol
x = 11.6508 M

3) Volume of 11.6508 M acid needed to deliver 0.1759032 mol:

0.1759032 mol / 11.6508 mol/L = 0.01509795 L

15.1 mL (to three sig figs) 

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Molarity Dafination and Explanation

Molarity Dafination and Explanation
                                                                                Molarity Problems

As should be clear from its name, molarity involves moles. Boy, does it!
The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.

This is probably easiest to explain with examples.

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Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.
What would be the molarity of this solution?

The answer is 1.00 mol/L. Notice that both the units of mol and L remain. Neither cancels.
A symbol for mol/L is often used. It is a capital M. So, writing 1.00 M for the answer is the correct way to do it.
Some textbooks make the M using italics and some put in a dash, like this: 1.00-M. When you handwrite it; a block capital M is just fine.
When you say it out loud, say this: "one point oh oh molar." You don't have to say the dash (if it's there). By the way, you sometimes see 1.00 M like this: 1.00-molar. A dash is usually used when you write the word 'molar.'
And never forget this: replace the M with mol/L when you do calculations. The M is the symbol for molarity, the mol/L is the unit used in calculations.

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Example #2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity?

The answer is 2.00 M.
Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of sucrose or sodium chloride or anything else contains the same number of chemical units. And that number is 6.022 x 10²³ units, called Avogadro's Number.

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Example #3: What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?

The answer is 0.300 M.

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Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.
Example #4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?
The solution to this problem involves two steps which will eventually be merged into one equation.

Step One: convert grams to moles.Step Two: divide moles by liters to get molality.

In the above problem, 58.44 grams/mol is the molar mass of NaCl. (There is the term "formula weight" and the term "molecular weight." There is a technical difference between them that isn't important right now. The term "molar mass" is a moe generic term.) To solve the problem:

Step One: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.Step Two: dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M).

Comment: remember that sometimes, a book will write out the word "molar," as in 0.500-molar.

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Example #5: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.

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Example #6: 80.0 grams of glucose (C₆H₁₂O₆, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?

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Notice how the phrase "of solution" keeps showing up. The molarity definition is based on the volume of the solution, NOT the volume of pure water used. For example, to say this:

"A one molar solution is prepared by adding one mole of solute to one liter of water."

is totally incorrect. It is "one liter of solution" not "one liter of water."
Be careful on this, especially when you get to molality.

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The most typical molarity problem looks like this:
What is the molarity of "whatever" grams of "whatever" substance dissolved in "whatever" mL of solution.
To solve it, you convert grams to moles, then divide by the volume, like this:

The two steps just mentioned can be combined into one equation. Notice that moles is part of both equations, so one equation can be substituted into the other. Let's substitute the first into the second and rearrange just a bit to get this:

The M stands for molarity, the V for volume. In the second question, GMW has been substituted for molar mass. GMW stands for gram-molecular weight, which is a very common synonym for molar mass.

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Example #7: When 2.00 grams of KMnO₄ (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results?

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This next example is the most common type you'll see:
Example #8: How many grams of KMnO₄ are needed to make 500.0 mL of a 0.200 M solution?

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Example #9: 10.0 g of acetic acid (CH₃COOH) is dissolved in 500.0 mL of solution. What molarity results?

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Example #10: How many mL of solution will result when 15.0 g of H₂SO₄ is dissolved to make a 0.200 M solution?

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Molality Problems

Molality Problems

Problem #1: A solution of H₂SO₄ with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?
Solution:

8.010 m means 8.010 mol / 1 kg of solvent8.010 mol times 98.0768 g/mol = 785.6 g of solute
785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution.
1785.6 g divided by 1.354 g/mL = 1318.76 mL
8.01 moles / 1.31876 L = 6.0739 M
6.074 M (to four sig figs)

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Problem #2: A sulfuric acid solution containing 571.4 g of H₂SO₄ per liter of solution has a density of 1.329 g/cm³. Calculate the molality of H₂SO₄ in this solution
Solution:

1 L of solution = 1000 mL = 1000 cm³1.329 g/cm³ times 1000 cm³ = 1329 g (the mass of the entire solution)
1329 g minus 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution)
571.4 g / 98.0768 g/mol = 5.826 mol of H₂SO₄
5.826 mol / 0.7576 kg = 7.690 m

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Problem #3: An aqueous solution is prepared by diluting 3.30 mL acetone (d = 0.789 g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity, molality and mole fraction of acetone in this solution?
Solution:
1) Preliminary calculations:

mass of acetone: (3.30 mL) (0.789 g/mL) = 2.6037 g
moles of acetone: 2.6037 g / 58.0794 g/mol = 0.04483 mol <--- need to look up formula of acetone
mass of solution: (75.0 mL) (0.993 g/mL) = 74.475 g
mass of water in the solution: 74.475 g - 2.6037 g = 71.8713 g
moles of water: 71.8713 g / 18.015 g/mol = 3.9896 mol

2) Molarity:

0.04483 mol / 0.0750 L = 0.598 M

3) Molality:

0.04483 mol / 0.0718713 kg = 0.624 m

4) Mole fraction:

0.04483 mol / (0.04483 mol + 3.9896 mol) = 0.0111

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Problem #4: Calculate the molality of 15.00 M HCl with a density of 1.0745 g/cm³
Solution:
1) Let us assume 1000. mL of solution are on hand. In that liter of 15-molar solution, there are:

15.00 mol/L times 1.000 L = 15.00 mole of HCl15.00 mol times 36.4609 g/mol = 546.9135 g of HCl

2) Use the density to get mass of solution

1000. mL times 1.0745 g/cm³ = 1074.5 g of solution1074.5 g minus 546.9135 g = 527.5865 g of water = 0.5275865 kg

3) Calculate molality:

15.00 mol / 0.5275865 kg = 28.43 m (to four sig figs)

Note: the mole fractions of water and HCl can also be calculated with the above data. There are 29.286 moles of water and 15.00 moles of HCl. You may work out the mole fractions on your own.

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Problem #5: You are given 450.0 g of a 0.7500 molal solution of acetone dissolved in water. How many grams of acetone are in this amount of solution?
Solution:

0.7500 molal means 0.7500 mole of solute (the acetone) per 1000 g of watermass of acetone ---> 58.0794 g/mol times 0.7500 mol = 43.56 g
mass of solution ---> 1000 g + 43.56 g = 1043.56 g
43.56 is to 1043.56 as x is to 450
x = 18.78 g

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Problem #6: A 0.391 m solution of the solute hexane dissolved in the solvent benzene is available. Calculate the mass (g) of the solution that must be taken to obtain 247 g of hexane (C₆H₁₄).
Solution:

0.391 mol times 86.1766 g/mol = 33.6950 g33.6950 g + 1000 g = 1033.6950 g
In other words, every 1033.6950 g of 0.391 m solution delivers 33.6950 g of hexane
33.6950 is to 1033.6950 as 247 is to x
x = 7577.46446 g
to three sig figs, 7.58 kg of solution

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Problem #7: Calculate the mass of the solute C₆H₆ and the mass of the solvent tetrahydrofuran that should be added to prepare 1.63 kg of a solution that is 1.42 m.
Solution:

1.42 m means 1.42 mole of C₆H₆ in 1 kg of tetrahydrofuran1.42 mol times 78.1134 g/mol = 110.921 g
110.921 g + 1000 g = 1110.921 g
110.921 is to 1110.921 as x is to 1630
x = 162.75 g
To check, do this:
162.75 g / 78.1134 g/mol = 2.08351 mol
1630 g - 162.75 g = 1467.25 g
2.08351 mol / 1.46725 kg = 1.42 m

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Problem #8: What is the molality of NaCl in an aqueous solution in which the mole fraction of NaCl is 0.100?
Solution:

A mole fraction of 0.100 for NaCl means the mole fraction of water is 0.900.Let us assume a solution is present made up of 0.100 mole of NaCl and 0.900 mole of water.
mass of water present ---> 0.900 mol times 18.015 g/mol = 16.2135 g
molality of solution ---> 0.100 mol / 0.0162135 kg = 6.1677 m
to three sig figs, 6.17 m

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Problem #9: Calculate the molality (m) of a 7.55 kg sample of a solution of the solute CH₂Cl₂ (molar mass = 84.93 g/mol) dissolved in the solvent acetone (CH₃COH₃C) if the sample contains 929 g of methylene chloride
Solution:

mass solvent ---> 7550 g - 929 g = 6621 g = 6.621 kgmoles solute ---> 929 g/ 84.93 g/mol = 10.9384 mol
molality = 10.9384 mol / 6.621 kg = 1.65 m

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Problem #10: What is the molality of a 3.75 M H₂SO₄ solution with a density of 1.230 g/mL?
Solution:
1) Determine mass of 1.00 L of solution:

1000 mL x 1.230 g/mL = 1230 g

2) Determine mass of 3.75 mol of H₂SO₄:

3.75 mol x 98.0768 g/mol = 367.788 g

3) Determine mass of solvent:

1230 - 367.788 = 862.212 g

4) Determine molality:

3.75 mol / 0.862212 kg = 4.35 molal (to three sig figs)

Problem #11: What is the molality of NaCl in an aqueous solution which is 4.20 molar? The density of the solution is 1.05 x 10³ g/L.
Solution:
1) Assume 1.00 L of the 4.20 M solution is present:

1.00 L of this solution contains 4.20 mole of NaCl.1.00 L times 1050 g/L = 1050 g of solution.

2) Determine mass of water in 1050 g of solution:

4.20 mol times 58.443 g/mol = 245.4606 g <--- mass of NaCl in solution1050 g - 245.4606 g = 804.5394 g

3) Calculate the molality:

4.20 mol / 0.8045394 kg = 5.22 m (to three sig figs)

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Problem #12: Calculate the molarity of a 3.58 m aqueous RbCl solution with a density of 1.12 g/mL.
Solution:
1) 3.58 m means this:

3.58 mole of RbCl in 1000 g of water.

2) Determine total mass of solution:

3.58 mol times 120.921 g/mol = 432.89718 g1000 g + 432.89718 g = 1432.89718 g

3) Determine volume of solution:

1432.89718 g / 1.12 g/mL = 1279.37 mL

4) Determine molarity:

3.58 mol / 1.27937 L = 2.80 M

Here's another problem of this type.

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Problem #13: Calculate the molality of a solution containing 16.5 g of naphthalene (C₁₀H₈) in 54.3 g benzene (C₆H₆).
Solution:

molality = moles of naphthalene / kilograms of benzene(16.5 g / 128.1732 g/mol) / 0.0543 kg = 2.37 m

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Problem #14: What is the molality of a solution consisting of 1.34 mL of carbon tetrachloride (CCl₄, density= 1.59 g/mL) in 65.0 mL of methylene chloride (CH₂Cl₂, density = 1.33 g/mL)?
Solution:
1) Moles CCl₄:

1.34 mL times 1.59 g/mL = 2.1306 g2.1306 g / 153.823 g/mol = 0.013851 mol

2) Mass of the methylene chloride:

65.0 mL times 1.33 g/mL = 86.45 g = 0.08645 kg

3) Molality:

0.013851 mol / 0.08645 kg = 0.160 m (to three sig figs)

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Problem #15: Determine concentration of a solution that contains 825 mg of Na₂HPO₄ dissolved in 450.0 mL of water in (a) molarity, (b) molality, (c) mole fraction, (d) mass %, and (e) ppm. Assume the density of the solution is the same as water (1.00 g/mL). Assume no volume change upon the addition of the solute.
Solution:
1) Molarity:

MV = mass / molar mass(x) (0.4500 L) = 0.825 g / 141.9579 g/mol
x = 0.0129 M

2) Molality:

0.825 g / 141.9579 g/mol = 0.00581158 mol0.00581158 mol / 0.4500 kg = 0.0129 m

3) Mole fraction:

Na₂HPO₄ ---> 0.825 g / 141.9579 g/mol = 0.00581158 mol
H₂O ---> 450.0 g / 18.015 g/mol = 24.97918401 molmole fraction of the water ---> 24.97918401 mol / (24.97918401 mol + 0.00581158 mol) = 0.9998

4) Mass percent:

water ---> (450 g / 450.825 g) * 100 = 99.8%

5) ppm:

ppm means the number of grams of solute per 1,000,000 grams of solution0.825 is to 450.825 as x is to 1,000,000

x = 1830 ppm 

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