Molarity

 Molarity

The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.

		moles of solute
Molarity	=	––––––––––––––
		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 (its mass is 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?

Solution:

		1.00 mol
Molarity	=	–––––––
		1.00 L

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?

Solution:

		2.00 mol
Molarity	=	–––––––
		1.00 L

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 0.750 mol is dissolved in 2.50 L of solution?

Solution:

		0.750 mol
Molarity	=	–––––––––	=  0.300 M
		2.50 L

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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.

By the way, here's another point: solutions are always considered to be fully-mixed. The solute has been completely dispersed throughout the entire extent of the solution. All samples from a well-mixed solution will show the same concentration when analyzed.

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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?

Solution:

There two steps to the solution of this problem. Eventually, the two steps will 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.

Solution:

1) Convert grams to moles:

25.0 g
––––––––––	=	0.210 mol
119.9 g/mol

2) Calculate the molarity:

0.210 mol
––––––––	=	0.280 M
0.750 L

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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?

Solution:

1) Convert grams to moles:

80.0 g
––––––––––	=	0.444 mol
180.0 g/mol

2) Calculate the molarity:

0.444 mol
––––––––	=	0.444 M
1.00 L

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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 should be "one liter of solution" not "one liter of water."

This is correct:

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

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The most typical molarity problem looks like this:

What is the molarity of _____ grams of [a chemical] dissolved in _____ mL (or L) of solution.

To solve it, you convert mass (in grams) to moles, then divide by the volume, like this:

mass
–––––––––	=	moles
molar mass

moles
––––––	=	molarity
volume

The two steps just mentioned can be combined into one equation. First, I'll rearrange the second equation (with M for molarity and V for volume):

moles = MV

Since the moles of the first equation equals the moles of the second equation, I can write this:

		mass
MV	=	–––––––––	<--- this is a very useful equation. Remember it!!!
		molar mass

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Example #7: When 2.50 grams of KMnO₄ (molar mass = 158.0 g/mol) is dissolved into 100. mL of solution, what molarity results?

		2.50 g
(x) (0.100 L)	=	––––––––––
		158.0 g/mol

x = 0.158 M (to three sig figs)

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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. mL of a 0.200 M solution?

		x
(0.200 mol/L) (0.500 L)	=	––––––––––
		158.0 g/mol

x = 15.8 g (to three sig figs)

Note the use of mol/L. In the actual calculation, use mol/L rather than M. The M is the symbol for molarity and is used in discussing molarity. However, the unit g/mol is what is actually used in a calculation.

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Example #9: 10.0 g of acetic acid (CH₃COOH) is dissolved in 500. mL of solution. What molarity results?

		10.0 g
(x) (0.500 L)	=	––––––––––
		60.05 g/mol

x = 0.333 M (to three sig figs)

Make sure to always use liters with this equation. Never mL or cm³ or anything else. Only liters. Or, if you prefer, dm³.

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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?

		15.0 g
(0.200 mol/L) (x)	=	––––––––––
		98.07 g/mol

x = 0.765 L (to three sig figs)

In the requested volume, 76.5 mL.

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Bonus Example: Carbon tetrachloride (CCl₄) has a density of 1.59 kg/L. What is the concentration (in mol/L) of pure CCl₄?

Solution:

Assume 1.00 L of CCl₄ is present.

(1.59 kg/L) (1.00 L) = 1.59 kg

(1.59 kg) (1000 g/kg) = 1590 g

1590 g / 153.823 g/mol = 10.336556 mol

10.336556 mol / 1.00 L = 10.3 M (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 / x

x = 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¯¹

x = (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¯¹

x = 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.

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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 mL

Cross-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. mL

Cross-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 mol

0.00300 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 mol

0.00625 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 react

0.08554459 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 M

0.05279 L is the sum of the two solution volumes.

4) Determine the sodium ion concentration:

0.08554459 mol / 0.05279 L = 1.620 M

The sole source of sodium ion is from the NaOH.

5) Determine the sulfate ion concentration:

0.02596011 mol / 0.05279 L = 0.4918 M

The 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 mL

6.44 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 mass

16.6667 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 g

H₂O ---> 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 HCl

0.0892299 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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Problem #31: You need to make an 80.0 g mixture of ethanol and water containing equal molar amounts of both, what mass of each substance would be required?

Solution:

1) Some preliminary information and discussion:

Ethanol has a molar mass of 46.0684 g/mol

Water has a molar mass of 18.0152 g/mol

Moles of ethanol = moles of water = x

(x) (46.0684) = mass of ethanol in the 80.0 g total

(x) (18.0152) = mass of water in the 80.0 g total

2) Solving for x:

(x) (46.0684) + (x) (18.0152) = 80.0

(x) (46.0684 + 18.015) = 80.0

(x) (64.0836) = 80.0

x = 80.0 / 64.0836

x = 1.24837 mol

1.25 moles of each. to three sig figs

3) Doing this for the second step works too:

46.0684x + 18.015x = 80.0

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Problem #32: What mass of pure sulfuric acid must be made up to 250 cm³ of aqueous solution so that the resulting solution has the same concentration as a potassium hydroxide solution containing 2.00 g of the pure alkali in 100 cm³?

Solution:

1) Determine the concentration of the KOH solution:

MV = mass / molar mass

(x) (0.1 L) = 2.00 g / 56.10564 g/mol

x = 0.35647 M

2) Determine the mass of sulfuric acid needed to make a 0.35647 M solution:

MV = mass / molar mass

(0.35647 mol/L) (0.25 L) = y / 98.0791 g/mol

y = 8.74 g

3) This problem was answered on an "answers" website like this:

(2.00 g KOH) / (56.10564 g KOH/mol) / (100 cm³) x (250 cm³) x (98.0791 g H₂SO₄/mol) = 8.74 g H₂SO₄

No additional comments were made. The above is a technique known as dimensional analysis. It is presented as one line of calculations and, typically, has no explanation added. The above answer inspired the follow comments.

4) You must recognize that the value connecting the KOH and H₂SO₄ solutions is that they have the same molarity. Here is the KOH set up:

	mass1
MV1 =	–––––
	MM1	<--- MM means molar mass

	mass1
M =	–––––––––––
	(MM1) (V1)

5) Here is the H₂SO₄ set up:

	mass2	<--- mass2 is the answer to the question
MV2 =	–––––
	MM2

	mass2
M =	–––––––––––
	(MM2) (V2)

6) Since M = M, we have this:

mass1		mass2
–––––––––––	=	–––––––––––
(MM1) (V1)		(MM2) (V2)

7) Cross-multiply:

(mass₁) (MM₂) (V₂) = (mass₂) (MM₁) (V₁)

8) Solve for mass₂:

	(mass1) (MM2) (V2)
mass2 =	–––––––––––––––––
	(MM1) (V1)

9) Insert numbers and solve:

	(2.00 g) (98.0791 g/mol) (250 cm3)
mass2 =	–––––––––––––––––––––––––––––
	(56.10564 g/mol) (100 cm3)

Up above, I converted cm³ to L, but I did not have to. In retrospect, that can be seen in the set up just above. If you compare the set up just above to the dimensional analysis set up, you will see that they are comparable.

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Problem #33: A solution is prepared by dissolving 19.20 g ammonium sulfate in enough water to make 117.0 mL of stock solution. A 12.00 mL sample of this stock solution is added to 59.20 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Solution:

1) Determine the molarity of the stock solution:

MV = mass / molar nass

(x) (0.1170 L) = 19.20 g / 132.1382 g/mol

x = 1.2419 M

2) Determine molarity of diluted solution (assume volumes are additive):

M₁V₁ = M₂V₂

(1.2419 mol/L) (12.00 mL) = (x) (71.20 mL) x = 0.209 M

3) Determine concentration of the ions:

(NH₄)₂SO₄(aq) ---> 2NH₄⁺(aq) + SO₄²¯(aq)

For every one mole of ammonium sulfate that dissolves, two moles of ammonium ion are present in solution as well as one mole of sulfate ions.

Therefore:

[NH₄⁺] = 0.418 M

[SO₄²¯] = 0.209 M

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Problem #34: A solution was prepared by adding pure KCl to deionized water. When the water in 1.00 mL of the solution was evaporated completely, the residue remaining weighed 90.0 mg. What was the molarity of the KCl solution.

Solution:

MV = mass / molar mass

(x) (1.00 mL) = 90.0 mg / 74.551 mg/mmole

(x) (1.00 mL) = 1.21 mmol

x = 1.21 M (to three sig figs)

Note the use of milligrams, milliliters and millimoles.

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Problem #35: An aqueous solution is 0.556 M in NaCl and 0.245 M in KCl. If 100.0 mL of the solution is left uncovered and all of the water evaporates away, calculate the mass of the residue.

Solution #1:

1) Moles of each solute:

NaCl ---> (0.556 mol/L) (0.1000 L) = 0.0556 mol

KCl ---> (0.245 mol/L) (0.1000 L) = 0.0245 mol

2) Mass of each solute:

NaCl ---> (0.0556 mol) (58.443 g/mol) = 3.24943 g

KCl ---> (0.0245 mol) (74.551 g/mol) = 1.82650 g

3) Add and round off:

3.24943 g + 1.82650 g = 5.07593

Use the 'rounding by 5' rule to get the final answer of 5.08 g

Solution #2: 1) Use MV = mass/molar mass to solve for the mass of NaCl.

2) Use MV = mass/molar mass to solve for the mass of KCl.

3) Add the two masses and round off to three sig figs.

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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 M

0.0097724 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)