Library Guides: Chemistry Textbook: Self-Assessments

1. Matter and MeasurementsToggle Dropdown
2. Atoms, Molecules and IonsToggle Dropdown
3. Composition of Substances and Solutions
4. Stoichiometry of Chemical ReactionsToggle Dropdown
5. ThermochemistryToggle Dropdown
6. GasesToggle Dropdown
7. Chemical Bonding and Molecular GeometryToggle Dropdown
8. Liquids, Solids, and Modern MaterialsToggle Dropdown
9. Solutions and Colligative PropertiesToggle Dropdown
10. KineticsToggle Dropdown
11. Chemical Equilibria and ApplicationsToggle Dropdown
12. ThermodynamicsToggle Dropdown
13. ElectrochemistryToggle Dropdown
14. AppendicesToggle Dropdown

Self-Assessments

Mole Concept 1
Mole Concept 2- atoms in a molecule
Percent composition and empirical formulas
Molarity and dilution
Ion concentration

How many moles of propylene are in 25.0 g of propylene (C₃H₆)?

Choose 1 answer

0.329

1.68

0.594

1.01 × 10²⁴

0.412

Moles and mass are related by molar mass. Use molar mass as the conversion factor

grams → moles; Conversion factor: 1mol = 42.08 g.
Calculation: 25.0g × (1 mol / 42.08 g) = 0.598 mol. Always check the units in the calculation. You can see that grams cancel to give moles.
Check: The molar mass of propylene tells us that 42.08 g is the mass of 1 mole of propylene. Therefore 25 grams should be less than a mole.

Moles and mass are related by molar mass. Use molar mass as the conversion factor.

How many moles of urea are in 5.4 x 10¹² molecules of urea, CO(NH₂)₂?

Choose 1 answer

9.0 × 10³⁴

9.0 × 10^-12

9.0 × 10¹⁰

1.1 × 10^-11

1.5 × 10^-13

Check your calculation. You may not have entered the calculation correctly in the calculator. All values multiplied together in the denominator must be enclosed in a single parenthesis.
4 x 10¹² molecules is less than 1 mole!

molecules → moles; Conversion factor: 1 mol = 6.022 ×10²³
Calculation: 5.4 x 10¹² molecules × (1 mol / 6.022 ×10²³ molecules) = 9.0 x 10^-12 after rounding to two significant digits.
Note that when inputting the calculation as a single line in a calculator, a parenthesis must be used as follows: 5.4 x 10¹² ÷ (6.022 × 10²³)

A collection of 6.022 × 10²³ urea molecules is 1 mole of urea.
4 x 10¹² molecules should be less than 1 mole!

A collection of 6.022 × 10²³ urea molecules is 1 mole of urea. Use this information as a conversion factor.

What is the mass of 2 water molecules?

Choose 1 answer

8 × 10^-21 g

1 × 10^-21 g

7 × 10^-22 g

2 × 10²⁵ g

6 × 10^-23 g

Convert number of molecules to grams via moles. molecules → moles → grams.
Moles and molecules are related by the Avogadro’s number. Moles and mass in grams are related by the molar mass.

You may have inverted the conversion factor between moles and molecules.
1 mol = 6.022 × 10²³ Write the conversion factor so that units in the denominator cancel out the units in the numerator.

molecules → moles → grams.
2 molecules × ( 1 mol / 022×10²³ molecules) × (18.02 g/ 1 mol) = 6×10^-23 g after rounding to one significant digit.
You can see that the units in the numerator and denominator cancel to give grams as the final unit.

Treflan (C₁₃H₁₆N₂O₄F) is a herbicide. Determine the number of molecules of Treflan in 25 lb of Treflan.

Choose 1 answer

2.4 × 10²⁵

3.6 × 10²³

5.3 × 10²²

9.3 × 10²⁴

1.6 × 10²⁵

lb → grams → moles → molecules. Write conversion factors so that units in the numerator and denominator cancel.
25 lb × (454 g/ 1 lb ) × (1 mol / 283.28 g) × (6.022×10²³ molecules / 1 mol) = 2.4 ×10²⁵.

Convert the given mass to grams. Then grams can be converted to molecules via moles

You may not have converted lb to grams.

You may not have used parenthesis while inputting the numbers into the calculator. Note that all values multiplied together in the denominator must be enclosed in a single parenthesis.

You may have inverted the conversion factor between grams and moles.
1 mol = 283.28 g. Write the conversion factor so that units in the denominator cancel out the units in the numerator.

What is the mass of 5.60 × 10²² molecules of glucose (C₆H₁₂O₆)?

Choose 1 answer

16.8 g

51.6 g

5.16 × 10^-4 g

1.68 × 10²⁵ g

A collection of 5.60× 10²² molecules is 0.093 mol, which corresponds to 16.8 g.
Strategy: 5.60 × 10²² molecules → moles → grams Conversion factors: 1 mol = 6.022 ×10²³ molecules; 1 mol glucose = 180.16 g. Conversion factors are written so that units in the denominator cancel out the units in the numerator

molecules → moles → grams.
To find the mass of 5.60 × 10²² molecules, first find how many moles 5.60 × 10²² molecules corresponds to. We know that 022×10²³ molecules correspond to 1 mole.

molecules → moles → grams.

To find the mass of 5.60 × 10²² molecules, first find how many moles 5.60 × 10²² molecules corresponds to. We know that 6.022×10²³ molecules correspond to 1 mole.

You may have inverted the conversion factor between grams and moles.
1 mol of glucose = 180.16 g. Write the conversion factor so that units in the denominator cancel out the units in the numerator.

You may have thought that what is given is moles. Note that what is given is molecules.
60 × 10²² molecules is less than one mole. So, the mass should be less than the molar mass

How many moles of sodium are in 5.4 g of Na₂SO₄?

Choose 1 answer

0.038 mol

0.076 mol

0.019 mol

0.12 mol

0.24 mol

What you have calculated is moles of Na₂SO₄. However, what is asked is moles of Na. Convert moles of Na₂SO₄ to moles of Na using the molar ratio between Na₂SO₄ and Na.

Strategy:4 g Na₂SO₄ → mol Na₂SO₄ → mol of Na
Conversion factors: Use the molar mass of Na₂SO₄ to convert grams to moles. Then, use the molar ratio between Na₂SO₄ and Na (since 1 Na₂SO₄ unit has 2 Na atoms, 1 mole Na₂SO₄ units will have 2 mole Na atoms. The molar ratio is 1 mol Na₂SO₄ = 2 mol Na)
Calculation: 5.4 g Na₂SO₄ × (1 mol Na₂SO₄ / 142.04 g Na₂SO₄) × ( 2 mol Na / 1 mol Na₂SO₄).
Notice that the units in the denominator cancel out the units in the numerator to yield mol of Na as the final unit.

You may have inverted the molar ratio between Na₂SO₄ and Na.
1 mol Na₂SO₄ = 2 mol Na. Write conversion factor so that the unit in the denominator cancels the unit in the numerator.

You may have used mass of sodium in the denominator! Note that what you are given is 5.4 g of Na₂SO₄. To cancel grams of Na₂SO₄conversion factor should have grams of Na₂SO₄ in the denominator.
Hint: The problem involves two conversion factors: molar mass of Na₂SO₄, molar ratio between Na₂SO₄ and Na.

You may have used molar mass of sodium in the denominator! Note that what you are given is 5.4 g of Na₂SO₄. To cancel grams of Na₂SO₄, the first conversion factor should have grams of Na₂SO₄in the denominator.
Hint: The problem involves two conversion factors: molar mass of Na₂SO₄, molar ratio between Na₂SO₄ and Na.

How many atoms of oxygen are in 1.5 g of Ca(NO₃)₂?

Choose 1 answer

1.9 × 10²³

3.3 × 10²²

9.2 × 10^>22

5.5× 10²¹

9.1 × 10^-26

You may have counted oxygen atoms twice!

Strategy: g of Ca(NO₃)₂ → mol Ca(NO₃)₂ → mol of O → atoms of O
Conversion factor for mol Ca(NO₃)₂ → mol of O is 1 mol Ca(NO₃)₂ = 6 mol O (1:6 molar ratio)

OR,

g of Ca(NO₃)₂ → mol Ca(NO₃)₂ → molecules of Ca(NO₃)₂ → atoms of O
Conversion factor for molecules of Ca(NO₃)₂ → atoms of O is 1 Ca(NO₃) molecule = 6 oxygen atoms (1:6 ratio).
Note: ‘formula unit’ is the correct terminology instead of ‘molecule’ when referring to ionic compounds. For simplicity, the word molecule was used here.

Review similar problem in worked out example 3.8

You may have used mass of oxygen in the denominator! Note that what you are given is 1.5 g of Ca(NO₃)₂. To cancel grams of Ca(NO₃)₂, the first conversion factor should have grams of Ca(NO₃)₂ in the denominator.
Hint: If you know how many Ca(NO₃)₂ units there are in 1.5 g Ca(NO₃)₂, you can count the oxygen atoms they contain-- knowing that each Ca(NO₃)₂ unit contains 6 oxygens.
Review similar problem in worked out example 3.8

5× 10²¹ is how many Ca(NO₃)₂ units there are, NOT how many oxygens! Note that each Ca(NO₃)₂ unit has 6 oxygens

You may have inverted the conversion factor involving Avogadro’s number. Note that 9.1×10^-26 has a negative exponent, so, this value is less than 1! The sample 1.5 g sample should contain lots of oxygen atoms.

How many atoms of chlorine are in 10.5 g of carbon tetrachloride?

Choose 1 answer

3.89 × 10²⁷

9.73 × 10²⁶

1.64 × 10²³

4.11 × 10²²

1.13 × 10²¹

Sorry, 3.89 × 10²⁷ is not the correct answer! The formula for carbon tetrachloride is CCl₄. Review similar problem in worked out example 3.8.

Sorry, 9.73 × 10²⁶is not the correct answer! The formula for carbon tetrachloride is CCl₄. Review similar problem in worked out example 3.8.

Strategy:
g of CCl₄ → mol of CCl₄ → molecules of CCl₄ → atoms of Cl
(Note: 1:4 ratio is used in the third conversion factor).
OR,
g of CCl₄ → mol of CCl₄→ mol of Cl → molecules of Cl
(Note: 1:4 molar ratio is used in the second conversion factor).

The formula for carbon tetrachloride is CCl₄. You may not have considered that one CCl₄ molecule has 4 chlorine atoms. Review similar problem in worked out example 3.8.

Sorry, 1.13 × 10²¹is not the correct answer! The formula for carbon tetrachloride is CCl₄. Review similar problem in worked out example 3.8.

How many ions are in 5.4 moles of Al₂(SO₄)₃?

Choose 1 answer

5.5 × 10²⁵

3.3 × 10²⁴

6.4 × 10²⁴

1.6 × 10²⁵

9.7 × 10²⁴

Error in counting ions; you may have counted 17 in each formula unit! Note that there are two aluminum ions and three sulfate ions, that is 2+3=5 ions, in each Al₂(SO₄)₂ unit (note that sulfate is a polyatomic ion, don’ dissociate it into sulfurs and oxygens!)

What you calculated (ie. 3.3 × 10²⁴) is how many Al₂(SO₄)₃ units there are. You did not consider that each Al₂(SO₄)₃ is made of positive and negative ions. Report how many ions there would be in 3.3 × 10²⁴ Al₂(SO₄)₃ units.

You may have counted only Aluminum ions. Count both Aluminum and sulfate ions. Note that there are two aluminum ions and three sulfate ions, that is 2+3=5 ions, in each Al₂(SO₄)₂ units.

There are 5 ions in each Al₂(SO₄)₃ unit, ie. 1:5 ratio between Al₂(SO₄)₃ and ions.
Strategy: Step-1: 4 moles Al₂(SO₄)₃→ number of molecules of Al₂(SO₄)₃using Avogadro’s number as conversion factor; Step-2: molecules of Al₂(SO₄)₃→ number of ions using 1:5 ratio.

OR,
Step-1: 5.4 moles Al₂(SO₄)₃→ mol of ions using 1:5 molar ratio; Step-2: mol of ions → number of ions using Avogadro’s number.

You may have counted only sulfate ions. Count both Aluminum and sulfate ions. Note that there are two aluminum ions and three sulfate ions, that is 2+3=5 ions, in each Al₂(SO₄)₂

Which of these contains the greatest mass of oxygen?

Choose 1 answer

2 mol of ethanol (C₂H₅OH)

1 mol of formic acid (HCO₂H)

1 mol of water (H₂O)

1 mol of SO₃

1 mol of CO₂

2 mol of C₂H₅OH contains 2 mol of oxygen (O) equivalent to 32 grams. There is another substance in the options that contains more than 32 grams.

1 mol of HCO₂H contains 2 mol of oxygen (O) equivalent to 32 grams. There is another substance in the options that contains more than 32 grams.

1 mol of H₂O contains 1 mol of oxygen (O) equivalent to 16 grams. There is another substance in the options that contains more than 16 grams.

1 mol of SO₃ contains 3 mol of oxygen (O) equivalent to 48 grams. Other substances in the options contain either 16 g or 32 g of oxygen.

1 mol of CO₂ contains 2 mol of oxygen (O) equivalent to 32 grams. There is another substance in the options that contains more than 32 grams.

What mass of NH₃ contains 5.60 g of hydrogen?

Choose 1 answer

94.6 g

31.5 g

28.8 g

18.5 g

10.1 g

The value, 94.6 g, is too much!
1 NH₃ has 3H. Interpret this in terms of mass and use the mass relationship as a conversion factor.

Strategy: 5.60 g of hydrogen (H) → grams of NH₃.
1 mole of NH₃ contains 3 moles of H. We can interpret this in terms of mass and say 17.02 g of NH₃ contains 3×008 g H (which equals 3.02 g ).
So, the conversion factor will be: 17.02 g NH₃ = 3.02 g H.

Calculation: 5.6 g H (17.02 g NH₃ / 3.02 g H) = 31.5 g H.
Note that the unit in the denominator cancels the unit in the numerator.

1 NH₃ has 3H. Interpret this in terms of mass and use the mass relationship as a conversion factor.

What is the percent composition of sulfur in a photographic fixer solution (“hypo”) that has the molecular formula, Na₂S₂O₃?

Choose 1 answer

10.1%

20.2%

28.6%

40.6%

6.73%

Percent composition of sulfur is percentage of sulfur “by mass” in the compound.
Percent is (part / whole)× 100, part being Sulfurs mass, whole being the compound’s mass.
1 mole Na₂S₂O₃ has 2 moles of S. Interpret this in terms of mass and calculate the percentage.

You may not have considered that there are two sulfur atoms in each Na₂S₂O₃ Both sulfur atoms contribute to the mass percentage of sulfur.

That is not the correct answer. You may have not considered that the percentage should be percentage by “mass”.
Percent is (part / whole)× 100, part being Sulfurs mass, whole being the compound’s mass.

Percent composition of sulfur is percentage of sulfur “by mass” in the compound.
Percent is (part / whole)× 100, part being Sulfurs mass, whole being the compound’s mass.
Mass percent of sulfur = ((2 x 32 g) /11 g) × 100

The value 6.73 is too small! Note that percent composition means percentage by “mass”.

What is the percent composition of calcium ion in Ca₃(PO₄)₂?

Choose 1 answer

51%

42%

39%

17%

13%

Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.

Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.
= ((3× 40.08 g) / 210.18 g) × 100 = 39%

(Note: Do not get confused because the problem states calcium “ion”. Ions weigh the same as atoms because removing or adding electrons has negligible effect on the mass of an atom. The problem should be solved as you would normally solve a percent composition problem.)

Percent by mass of calcium = (mass of Ca in the compound / mass of the compound) × 100.

You may not have considered that there are 3 calcium ions in each Ca₃(PO₄)₂. All the three calcium ions contribute to the mass percentage of calcium.

Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its empirical formula?

Choose 1 answer

C₆HCl₁₈

C₂H₂Cl₄

C₂H₄Cl₂

CH₂Cl

CH₂

You may have calculated simplest ratio of grams, which is not correct. Note that empirical formula is the simplest whole number ratio of moles!
Review similar worked out problem in example 3.12.

ote that C₂H₂Cl₄ can be simplified to CHCl₂. So, C₂H₂Cl₄ cannot be the empirical formula! Also, the ratio is not correct.
Review similar worked out problem in example 3.12.

It is true that the ratio of moles is 2:4:2. However, this is not the simplest ratio!
Empirical formula is the simplest whole number ratio of moles.
Review similar worked out problem in example 3.12.

Assuming 100 g of the compound:
Ratio of masses: 3 g C, 4.1 g H, 71.6 g Cl
Ratio of moles: 2.023 mol C, 4.067 mol H, 2.020 mol Cl
This results in a relative ratio of 1.001 : 2.013 : 1.000, which is close to whole number ratio, 1:2:1. Hence, the Empirical formula is CH₂Cl.
Review similar worked out problem in example 3.12.

Do not ignore chlorine in the compound! Since mass percentages of all components should add up to 100%, Chlorine’s mass percentage will be 71.6%.
Review similar worked out problem in example 3.12.

Chrysotile asbestos (MW: 520.8) has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. What are the empirical and molecular formulas of Chrysolite?
(MW refers to Molecular weight, 520.8 g/mol)

Choose 1 answer

Empirical: Mg₆Si₄H₂₂O₁₅, Molecular: Mg₆Si₄H₂₂O₁₅

Empirical: Mg₃Si₂H₃O₈,Molecular: Mg₆Si₄H₆O₁₆

Empirical: MgSiHO₃, Molecular: Mg₄Si₄H₄O₁₂

Empirical: Mg₂SiH₂O₄, Molecular: Mg₄Si₂H₄O₈

Empirical: Mg₂SiH₂O₃, Molecular: Mg₆Si₃H₆O₉

This is not the correct answer. If you have calculated the percentages from the formula, you can see that the method may not always work. Determine the empirical and molecular formulas systematically as shown in a similar worked out problem, example 3.12.

Grams: 28.03 g Mg, 21.60 g Si, 1.16 g H, 49.21 g O (assuming 100 g sample)
Moles: 1.17 mol Mg, 0.769 mol Si, 1.15 mol H, 3.08 g O. This results in a relative ratio of 1.5 : 1 : 1.5 : 4; doubling each results in simplest whole number ratio, 3:2:3:8. Therefore, empirical formula is Mg₃Si₂H₃O₈. The molar mass is twice the empirical formula mass. Hence, the molecular formula is Mg₆Si₄H₆O₁₆.
Review similar worked out problem in example 3.12.

Review similar worked out problem in example 3.12.

Unless simplest ratio is very close to a whole number ratio, do not round!
Review similar worked out problem in example 3.12.

Review similar worked out problem in example 3.12.

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas.
What is the empirical formula of a polymer that contains 40.0 g carbon and 6.71 g hydrogen and 53.28 g oxygen?

Choose 1 answer

C₃H₇O₃

C₆HO₈

C₄H₂O

C₃H₆O₃

CH₂O

Review similar worked out problem in example 3.12.

Review similar worked out .problem in example 3.12

Grams: 40.0 g C, 6.71 g H, 53.28 g O.
Moles: 3.331 mol C, 6.657 mol H, 3.330 mol O.
This results in a relative ratio of 1.000 : 1.999 : 000, which is close to simplest whole number ratio of 1:2:1. Therefore, empirical formula is CH₂O.
Review similar worked out problem in example 3.12.

How many milliliters of water are needed to prepare 250 mL of 0.025 M KNO₃ solution from a 0.150 M KNO₃ solution?

Choose 1 answer

210 mL

110 mL

98 mL

42 mL

32 mL

M₁V₁ = M₂V₂, where, V₁ and V₂ are volumes of solution before and after adding water, respectively. We know the volume after dilution to be 250 mL. Calculate the volume before dilution. The difference between the two volumes is how much water is needed.
15M × V₁ = 0.025M × 250 mL
V₁ = 41.7 mL.
V_water = V₂ – V₁ = 208 mL = 210 mL after rounding to two significant digits.

This is a dilution problem.
The solution is prepared from a 0.15 M That means, molarity of the solution before dilution, M₁, is 0.150M. The molarity and volume of the solution after dilution (i.e. M₂ and V₂) are 0.025M and 250 mL. Using this information, we can first calculate M₁ and find the volume of water.

Use the dilution equation, M₁V₁ = M₂V₂. Note that V₁ and V₂ are volumes of solution before and after adding water.

Looks like you stopped after calculating the volume of solution before dilution. The volume of water needed is given by the difference in the volumes before and after dilution.

Use the dilution equation, M₁V₁ = M₂V₂. Note that V₁ and V₂ are volumes of solution before and after adding water.

How many moles of KNO₃ is in 23.15 mL of 0.0120 M solution?

Choose 1 answer

2.78 × 10^-4mol

5.24 × 10^-4mol

0.278 mol

0.524 mol

1.93 mol

Rearranging the molarity equation gives: moles = molarity × Liters = 0.0120 M ×02315 L = 2.78 × 10^-4 mol.
Check calculation from units: Note that the unit, “M” stands for “mol/L”. You can see that in the above calculation, Liters cancel out and the resulting unit is mol.

See more example problems based on the molarity equation here (examples 3.14 to 3.18).

You did not rearrange the molarity equation correctly! Also, pay attention to units in your calculation.

You did not convert milliliters to Liters! Recall that Molarity = mol/L.

Check if you rearranged the molarity equation correctly!

A 100.0 mL of water is added to 25.0 mL of 0.500 M NaCl solution. What the new concentration of the solution?

Choose 1 answer

0.100 M

0.125 M

0.150 M

0.175 M

2.00 M

This is a dilution problem. As more solvent is added, concentration decreases. We can solve for the new concentration using the dilution equation is M₁V₁ = M₂V₂. The volume and molarity before dilution, V₁ and M₁, are 25 mL and 0.5 M. The volume after dilution, V₂, is 125 mL (since 100 mL water is added to 25 mL solution).

Calculation:500M × 25.0 mL = M₂ × 125 mL. M₂ = 0.500M × 25.0 mL / 125 mL = 0.100 M.
Note that volume increased by a factor of 5 (from 25 mL to 125 mL), so, concentration decreased by a factor of 5 (from 0.5 M to 0.1 M)

Sorry, 0.125M is too low! Review the concept here.
Check values for volume in your calculation! Volume of the solution after dilution will be 25 mL plus 100 mL.
Review the dilution concept here.

Use the dilution equation. Note that the volumes V₁ and V₂ refer to volume before and after dilution, respectively. Review the concept here!

Concentration should decrease after adding water! 2.00 M is a higher concentration than the starting concentration of 0.500 M. Hence, 2.00 M cannot be the answer.
Review the dilution concept here.

What volume of a 250 mL stock solution of 2.0 M HNO₃ is needed to prepare 150 mL of 0.041 M HNO₃ solution?(g)

Choose 1 answer

246 mL

73 mL

16 mL

3.1 mL

2.0 mL

Hint: 250 mL will not be used anywhere in the calculation. The value, 250 mL, is simply stating how much of 2.0 M HNO₃ solution is available.

M₁V₁ = M₂V2.0M × V₁ = 0.041M × 150 mL -----> V₁ = 3.1 mL. You will need 3.1 mL of 2.0 M solution.
Note: If water is added to the 3.1 mL until the volume is 150 mL, the new molarity of the solution will be 0.041 M.

Hint: 250 mL will not be used anywhere in the calculation. The value, 250 mL, is simply stating how much of 2.0 M HNO₃ solution is available.
The question can be simplified to “what volume of 2.0M solution is needed to prepare a 150 mL solution of 0.041M

How many milliliters of water are needed to prepare 250 mL of 0.025 M KNO₃ solution from a 0.150 M KNO₃ solution?

Choose 1 answer

210 mL

110 mL

98 mL

42 mL

32 mL

Use the dilution equation, M₁V₁ = M₂V₂. Note that V₁ and V₂ are volumes of solution before and after adding water.

Looks like you stopped after calculating the volume of solution before dilution. The volume of water needed is given by the difference in the volumes before and after dilution.

Use the dilution equation, M₁V₁ = M₂V₂. Note that V₁ and V₂ are volumes of solution before and after adding water.

A 2.1 gram sample of CaCl₂is dissolved in water to prepare a 150 mL solution. What is the molarity of chloride ions in the solution?

Choose 1 answer

0.10 M

0.20 M

0.13 M

0.25 M

0.028 M

Note that molarity is moles/L. Relate moles of CaCl₂to moles of Cl^-.

Sorry! 0.13 mol/L is the concentration of CaCl₂, not the concentration of Cl^--

Convert molarity of CaCl₂to molarity of Cl^-using the molar ratio between CaCl₂and Cl^-.

Good job! CaCl₂dissociates in water to produce one Ca²⁺and two Cl^- Note that the molar ratio between CaCl₂and Cl^-is 1:2.

1 grams of CaCl₂in 150 mL solution. We can calculate the concentration (ie. molarity) of CaCl₂using the molarity equation. Since each CaCl₂produces two chloride ions when it dissolves in water, concentration of Chloride (Cl^-) will be two times the concentration of CaCl₂.

1 g CaCl₂× (1 mol CaCl₂/ 110.98 g CaCl₂) = 0.0189 mol CaCl₂.

Molarity of CaCl₂= 0.0189 mol CaCl₂/ 0.15 L solution = 0.126 mol CaCl₂/ L

(0.126 mol CaCl₂/ L) × (2 mol Cl^-/ 1 mol CaCl₂) = 0.25 mol Cl^-/L

Note that molarity is moles/L. Relate moles of CaCl₂to moles of Cl^-.

How many moles of nitrate ions are in 20 mL of 0.15 M Mg(NO₃)₂solution?

Choose 1 answer

0.006 M

0.003 M

0.4 M

6 M

3 M

Molarity of Mg(NO₃)₂x Liters of solution = moles of Mg(NO₃)₂moles of Mg(NO₃)₂→ moles of NO₃^-using the molar ratio between Mg(NO₃)₂and NO₃^-

You did not convert moles of Mg(NO₃)₂to moles of NO₃^-

We can calculate moles of molarity and volume are given. Using the molar ratio between

Mg(NO₃)₂and NO₃^-we can convert moles of Mg(NO₃)₂to moles of NO₃^-.

Recall the molarity equation. Think about what is given and what you need to calculate.

How many grams of Aluminum bromide are needed to prepare a 250 mL solution with a bromide ion concentration of 0.050 M?

Choose 1 answer

0.65 g

1.1 g

3.3 g

4.2 g

8.4 g

Determine the formula for aluminum bromide. Note that we can convert bromide concentration (Br^-) to AlBr₃concentration using the molar ratio.

You did not convert moles of Br^-to moles of AlBr₃! Note that each AlBr₃contains 3 Br^- ions.

Aluminum bromide is AlBr₃.

Molarity of Br^-× Liters of solution= moles of Br^-.

moles of Br^- → moles of AlBr₃ (using the molar ratio: 1 mol AlBr₃ = 3 mol Br^- as the conversion factor).

mol AlBr₃ → grams of AlBr₃

Determine the formula for aluminum bromide. Note that we can convert bromide concentration (Br^-) to AlBr₃concentration using the molar ratio.

How many grams of sodium is in 100.0 mL of 0.20 M Na₂SO₄solution?

Choose 1 answer

2.8 g

0.92 g

0.70 g

0.46 g

0.17 g

Think about what you can calculate from Molarity of Na₂SO₄and it’s volume.

Molarity of Na₂SO₄× Liters of solution = moles of Na₂SO₄→ moles of Na → grams of Na

Using molarity of Na₂SO₄and the volume of the solution, we can find moles of Na₂SO₄. moles of Na₂SO₄can be converted to moles of sodium (Na)

Which of the following solutions contains the highest concentration of chloride ions?

Choose 1 answer

0.1 M NaCl

0.1 M CaCl₂

0.1 M AlCl₃

0.1 M NaClO₃

All of them have the same concentration of chloride

Chloride concentration in 0.1 M NaCl solution will also be 0.1 M.

Chloride concentration in 0.1 M CaCl₂solution will be 0.2 M.

1 mole AlCl₃contains 3 moles Cl^-. Due to this molar ratio, chloride concentration in 0.1 M AlCl₃solution will be 0.3 M.

NaClO₃has Na⁺and ClO₃^- It does not contain chloride (Cl^-) ions. Chloride concentration in NaClO₃solution is zero!

Though all solutions contain the same moles of solute per liter, each solute has different number of Chloride ions!

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Chemistry Textbook

Self-Assessments

How many moles of propylene are in 25.0 g of propylene (C3H6)?

How many moles of urea are in 5.4 x 1012 molecules of urea, CO(NH2)2?

What is the mass of 2 water molecules?

Treflan (C13H16N2O4F) is a herbicide. Determine the number of molecules of Treflan in 25 lb of Treflan.

What is the mass of 5.60 × 1022 molecules of glucose (C6H12O6)?

How many moles of sodium are in 5.4 g of Na2SO4?

How many atoms of oxygen are in 1.5 g of Ca(NO3)2?

How many atoms of chlorine are in 10.5 g of carbon tetrachloride?

How many ions are in 5.4 moles of Al2(SO4)3?

Which of these contains the greatest mass of oxygen?

What mass of NH3 contains 5.60 g of hydrogen?

What is the percent composition of sulfur in a photographic fixer solution (“hypo”) that has the molecular formula, Na2S2O3?

What is the percent composition of calcium ion in Ca3(PO4)2?

Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its empirical formula?

Chrysotile asbestos (MW: 520.8) has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. What are the empirical and molecular formulas of Chrysolite?(MW refers to Molecular weight, 520.8 g/mol)

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. What is the empirical formula of a polymer that contains 40.0 g carbon and 6.71 g hydrogen and 53.28 g oxygen?

How many milliliters of water are needed to prepare 250 mL of 0.025 M KNO3 solution from a 0.150 M KNO3 solution?

How many moles of KNO3 is in 23.15 mL of 0.0120 M solution?

A 100.0 mL of water is added to 25.0 mL of 0.500 M NaCl solution. What the new concentration of the solution?

What volume of a 250 mL stock solution of 2.0 M HNO3 is needed to prepare 150 mL of 0.041 M HNO3 solution?(g)

How many milliliters of water are needed to prepare 250 mL of 0.025 M KNO3 solution from a 0.150 M KNO3 solution?

A 2.1 gram sample of CaCl2 is dissolved in water to prepare a 150 mL solution. What is the molarity of chloride ions in the solution?

How many moles of nitrate ions are in 20 mL of 0.15 M Mg(NO3)2 solution?

How many grams of Aluminum bromide are needed to prepare a 250 mL solution with a bromide ion concentration of 0.050 M?

How many grams of sodium is in 100.0 mL of 0.20 M Na2SO4 solution?

Which of the following solutions contains the highest concentration of chloride ions?

How many moles of propylene are in 25.0 g of propylene (C₃H₆)?

How many moles of urea are in 5.4 x 10¹² molecules of urea, CO(NH₂)₂?

Treflan (C₁₃H₁₆N₂O₄F) is a herbicide. Determine the number of molecules of Treflan in 25 lb of Treflan.

What is the mass of 5.60 × 10²² molecules of glucose (C₆H₁₂O₆)?

How many moles of sodium are in 5.4 g of Na₂SO₄?

How many atoms of oxygen are in 1.5 g of Ca(NO₃)₂?

How many ions are in 5.4 moles of Al₂(SO₄)₃?

What mass of NH₃ contains 5.60 g of hydrogen?

What is the percent composition of sulfur in a photographic fixer solution (“hypo”) that has the molecular formula, Na₂S₂O₃?

What is the percent composition of calcium ion in Ca₃(PO₄)₂?

Chrysotile asbestos (MW: 520.8) has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. What are the empirical and molecular formulas of Chrysolite?
(MW refers to Molecular weight, 520.8 g/mol)

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas.
What is the empirical formula of a polymer that contains 40.0 g carbon and 6.71 g hydrogen and 53.28 g oxygen?

How many milliliters of water are needed to prepare 250 mL of 0.025 M KNO₃ solution from a 0.150 M KNO₃ solution?

How many moles of KNO₃ is in 23.15 mL of 0.0120 M solution?

What volume of a 250 mL stock solution of 2.0 M HNO₃ is needed to prepare 150 mL of 0.041 M HNO₃ solution?(g)

How many milliliters of water are needed to prepare 250 mL of 0.025 M KNO₃ solution from a 0.150 M KNO₃ solution?

A 2.1 gram sample of CaCl₂is dissolved in water to prepare a 150 mL solution. What is the molarity of chloride ions in the solution?

How many moles of nitrate ions are in 20 mL of 0.15 M Mg(NO₃)₂solution?

How many grams of sodium is in 100.0 mL of 0.20 M Na₂SO₄solution?