Call Now to Set Up Tutoring:

(888) 888-0446

AP Chemistry : Solutions and States of Matter

Study concepts, example questions & explanations for AP Chemistry

Create An Account

All AP Chemistry Resources

6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Concentration And Units

A chemist wants to turn a 50.0mL solution of $0.60 M\ HNO_{3}$ into a .40 M solution. How much water should she add?

Possible Answers:

25 mL

75 mL

13 mL

30 mL

65 mL

Correct answer:

25 mL

Explanation:

To solve this problem, we may use the following equation relating the molarity and volume of two solutions:

M_1V_1=M_2V_2

Recall:

$M=\frac{mol}{L}$

Plug in known values and solve.

(0.6M)(50mL)=(0.4M)(V_2)

V_2=75mL

However, this is not the final answer. The whole volume of the second, 0.4M solution is 85mL. Thus the chemist needs to add 25mL of water to the original solution to obtain the desired concentration.

Report an Error

Example Question #386 : Ap Chemistry

In order to dilute a 1mL solution that is 0.01M so that the solution is diluted to $5*10^{-3} M,$ , how many milliliters does this solution need to be diluted to?

Possible Answers:

20mL

1mL

2mL

10mL

5mL

Correct answer:

2mL

Explanation:

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{M_{2}}=V_{2}$

Plugging in the values gives:

$V_{2}=\frac{0.01\ M * 1\ mL}{5*10^{-3}\ M}= 2\ mL$

Report an Error

Example Question #383 : Ap Chemistry

What concentration would you have prepared if you diluted 30mL of a 0.350M salt solution to 50mL?

Possible Answers:

0.50M

0.75M

0.24M

0.89M

0.21M

Correct answer:

0.21M

Explanation:

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{V_{2}}=M_{2}$

Plugging in the values gives:

$M_{2}=\frac{0.35\ M * 30\ mL}{50\ mL}= 0.21 M$

Therefore, after diluting the solution to 50mL, the solution concentration would be lowered from 0.35M to 0.21M.

Report an Error

Example Question #32 : Solutions

How many moles are in a 0.010L solution with a concentration that is $5*10^{-3}M$ ?

Possible Answers:

$3*10^{-2}mol$

5.32mol

$2*10^{-5}mol$

$5*10^{-5}mol$

0.112mol

Correct answer:

$5*10^{-5}mol$

Explanation:

By using the concentration as a conversion factor, the number of moles can calculated by multiplying the concentration by the number of liters.

$\frac{5*10^{-3}moles}{L} * 0.010L = 5*10^{-5}moles$

Therefore, there are $5*10^{-5}moles$ in 0.010 L of a $5*10^{-3}M$ solution.

Report an Error

Example Question #31 : Solutions

In order to prepare a needed solution for an experiment, 0.082 grams of NaCl was dissolved in water to give a 35mL solution. What is the molarity of this solution?

Possible Answers:

0.40M

0.040M

0.34M

0.409M

0.45M

Correct answer:

0.040M

Explanation:

In order to calculate the concentration, we must use molarity formula:

$Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

We must use the molecular weight of sodium chloride to calculate the moles of solute:

$MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)$

$MW_{NaCl}=(23 \frac{g}{mole})+(35 \frac{g}{mole})= 58 \frac{g}{mole}$

$0.082g*\frac{mole}{58g}=0.0014\ moles\ NaCl$

$\frac{0.0014\ moles\ CH_{3}CH_{2}OH}{0.035\L} = 0.040\ M$

Therefore, the concentration in molarity of this solution is 0.040M.

Report an Error

Example Question #34 : Solutions

How many moles are in 1000mL solution with a concentration that is $3*10^{-5}M$ ?

Possible Answers:

$6.5*10^{-5}mol$

$3*10^{-8}mol$

$3*10^{-5}mol$

$9*10^{-5}mol$

$8*10^{-5}mol$

Correct answer:

$3*10^{-5}mol$

Explanation:

By simply using the concentration as a conversion factor, the number of moles can be calculated by multiplying the concentration by the number of liters. Before calculating the number of moles, the number of milliliters must be converted to liters using the fact that $1\ liter = 1000\ mL$ .

$\frac{3*10^{-5}\ moles}{L} * 1\ L = 3*10^{-5}\ moles$

Therefore, there are $3*10^{-5}\ moles$ in $0.010\ L$ of a $5*10^{-3}M$ solution.

Report an Error

Example Question #1 : Precipitates And Calculations

A chemist combines 300 mL of a 0.3 M $Na_2SO_4$ solution with 200 mL of 0.4 M $BaCl_2$ solution. How many grams of precipitate form?

Possible Answers:

233.4 g

58.5 g

18.7g

No precipitate is formed

21.0 g

Correct answer:

18.7g

Explanation:

First, let us write out an ion exchance reaction for the reactants:

$Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl\hspace{1 mm}+BaSO_4$

By solubility rules, $NaCl$ is soluble in water and $BaSO_4$ is not. Our new reaction is:

$Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl_{(aq)}\hspace{1 mm}+BaSO_{4\hspace{1 mm}(s)}$

Now we will calculate the theoretical yield of each reactant.

$300\hspace{1 mm}mL\times\frac{1 L}{1000 mL}\times \frac{0.3\hspace{1 mm}moles\hspace{1 mm}Na_2SO_4}{1 L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}Na_2SO_4}\times \frac{233.4\hspace{1 mm}g\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}=21.0\hspace{1 mm}g\hspace{1 mm}BaSO_4$

Now we perform the same calculation beginning with $BaCl_2$ :

$200\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.4\hspace{1 mm}moles\hspace{1 mm}BaCl_2}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaCl_2}\times\frac{233.4\hspace{1 mm}g\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}=18.7\hspace{1 mm}g\hspace{1 mm}BaSO_4$

The limiting reagent is $BaCl_2$ and this reaction produces 18.7 g precipitate.

Report an Error

Example Question #2 : Precipitates And Calculations

A chemist boils off the water from 234 mL of a 0.4 M solution of KCl. What is the mass of the remaining solid?

Possible Answers:

$234\hspace{1 mm}g\hspace{1 mm}KCl$

$6.98\hspace{1 mm}g\hspace{1 mm}KCl$

$29.82\hspace{1 mm}g\hspace{1 mm}KCl$

$74.55\hspace{1 mm}g\hspace{1 mm}KCl$

None of the available answers

Correct answer:

$6.98\hspace{1 mm}g\hspace{1 mm}KCl$

Explanation:

$234\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.4\hspace{1 mm}moles\hspace{1 mm}KCl}{1\hspace{1 mm}L}\times\frac{74.55\hspace{1 mm}g\hspace{1 mm}KCl}{1\hspace{1 mm}mole\hspace{1 mm}KCl}=6.98\hspace{1 mm}g\hspace{1 mm}KCl$

Report an Error

Example Question #2 : Precipitates And Calculations

A chemist boils off 344 mL of a 0.35 M solution of sodium hydroxide. How much solid remains?

Possible Answers:

$4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

No solid will remain as it will boil off with the water

$4.67\hspace{1 mm}g\hspace{1 mm}NaOH$

None of the available answers

$13.9\hspace{1 mm}g\hspace{1 mm}NaOH$

Correct answer:

$4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

Explanation:

First, you must recognize that the chemical formula for sodium hydroxide is NaOH . The mass of the boiled solution is

$344\hspace{1 mm}mL\times\frac{0.35\hspace{1 mm}mol\hspace{1 mm}NaOH}{1000\hspace{1 mm}mL}\times\frac{39.997\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mol\hspace{1 mm}NaOH}= 4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

Report an Error

Example Question #31 : Solutions

What is the mass of the solid left over after boiling off 100mL of 0.4M NaCl solution?

Possible Answers:

2.35g

0.0445g

2.34g

None of the available answers

0.04g

Correct answer:

2.34g

Explanation:

The remaining mass with be equal to the mass of the sodium chloride in the solution. Once the solvent (water) evaporates, the solute will remain.

Atomic mass of sodium is ~23. Atomic mass of chlorine is ~35.5. Molecular mass of NaCl is ~58.5.

$0.100L*\frac{0.4mol}{1L}=0.040mol\ NaCl$

$0.040mol\ NaCl*\frac{58.5g\ NaCl}{1mol\ NaCl}=2.34g\ NaCl$

Report an Error

View AP Chemistry Tutors

Melodi
Certified Tutor

Case Western Reserve University, Bachelor of Science, Computer Science.

View AP Chemistry Tutors

Brianna
Certified Tutor

Fairfield University, Bachelor of Science, Biology, General. National University of Ireland at Galway, Master of Science, Neu...

View AP Chemistry Tutors

Pedro
Certified Tutor

Harvard University, Bachelor of Chemistry, Chemistry.

All AP Chemistry Resources

6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Seattle, ACT Tutors in Atlanta, LSAT Tutors in Dallas Fort Worth, English Tutors in Washington DC, GRE Tutors in Miami, Algebra Tutors in Atlanta, GMAT Tutors in Boston, Computer Science Tutors in San Diego, Algebra Tutors in San Francisco-Bay Area, LSAT Tutors in Atlanta

Popular Courses & Classes

SAT Courses & Classes in New York City, SAT Courses & Classes in Miami, SSAT Courses & Classes in Houston, SSAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Miami, ACT Courses & Classes in Atlanta, GMAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in San Diego, LSAT Courses & Classes in Denver, ISEE Courses & Classes in San Diego

Popular Test Prep

SAT Test Prep in Seattle, SSAT Test Prep in New York City, LSAT Test Prep in Miami, SSAT Test Prep in Denver, SAT Test Prep in Dallas Fort Worth, GRE Test Prep in Dallas Fort Worth, ISEE Test Prep in Seattle, SAT Test Prep in Washington DC, GMAT Test Prep in San Diego, ISEE Test Prep in Miami

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Chemistry : Solutions and States of Matter

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #11 : Concentration And Units

Example Question #386 : Ap Chemistry

Example Question #383 : Ap Chemistry

Example Question #32 : Solutions

Example Question #31 : Solutions

Example Question #34 : Solutions

Example Question #1 : Precipitates And Calculations

Example Question #2 : Precipitates And Calculations

Example Question #2 : Precipitates And Calculations

Example Question #31 : Solutions

All AP Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: