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Example Questions

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Example Question #31 : Solutions

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

Possible Answers:

0.50M

0.24M

0.21M

0.89M

0.75M

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.

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Example Question #32 : Solutions

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

Possible Answers:

$5*10^{-5}mol$

5.32mol

$3*10^{-2}mol$

0.112mol

$2*10^{-5}mol$

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.

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Example Question #33 : 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:

0.409M

0.40M

0.45M

0.34M

0.040M

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.

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Example Question #34 : Solutions

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

Possible Answers:

$3*10^{-8}mol$

$9*10^{-5}mol$

$6.5*10^{-5}mol$

$3*10^{-5}mol$

$8*10^{-5}mol$

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.

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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:

18.7g

21.0 g

58.5 g

233.4 g

No precipitate is formed

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.

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Example Question #1 : 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:

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

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

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

None of the available answers

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

Correct answer:

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

Explanation:

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

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Example Question #3 : 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

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

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

None of the available answers

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

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

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Example Question #4 : Precipitates And Calculations

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

Possible Answers:

None of the available answers

2.34g

0.04g

0.0445g

2.35g

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.

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

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

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Example Question #1 : Precipitates And Calculations

A chemist has 5.2L of a 0.3M $Li_3 PO_4$ solution. If the solvent were boiled off, what would be the mass of the remaining solid?

Possible Answers:

$181g\ Li_3PO_4$

$108g\ Li_3PO_4$

$116g\ Li_3PO_4$

$53.9g\ Li_3PO_4$

$1.56g\ Li_3PO_4$

Correct answer:

$181g\ Li_3PO_4$

Explanation:

First we will figure out the number of moles of $Li_3 PO_4$ that we have. We have 5.2 L of a 0.3 M solution, so:

$0.3\hspace{1mm} \frac{mol}{L} \times 5.2\hspace{1 mm} L = 1.56 \hspace{1 mm}moles$

Now the problem is to find the mass of 1.56 moles of $Li_3 PO_4$ . First, we need to know the molecular weight of $Li_3 PO_4$ . We will go to the periodic table and add up the mass of each element present.

$(3\hspace{1 mm}\times6.94\hspace{1 mm}\frac{grams\hspace{1 mm} Li}{mol\hspace{1 mm}Li})+30.97\hspace{1 mm}\frac{grams\hspace{1 mm}P}{mol\hspace{1 mm}P}+(4\hspace{1 mm}\times 16\hspace{1 mm}\frac{grams\hspace{1 mm}O}{mol\hspace{1mm}O})= 115.79\hspace{1mm}\frac{grams\hspace{1 mm} Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}$

Now the problem is simple as we have the molar mass and the number of desired moles.

$115.79\hspace{1 mm}\frac{grams\hspace{1 mm}Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}\times1.56\hspace{1 mm}moles\hspace{1 mm}Li_3 PO_4 = 181\hspace{1 mm}grams\hspace{1 mm}Li_3 PO_4$

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Example Question #6 : Precipitates And Calculations

Given a pK_aof 6.37 for the first deprotonation of carbonic acid ( H_2CO_3 ), what is the ratio of bicarbonate ( HCO_3^- ) to carbonic acid ( H_2CO_3 ) at pH 5.60?

Assume that the effect of the deprotonation of bicarbonate is negligible in your calculations.

Possible Answers:

$1.07 * 10^{-12}$

5.89

0.170

$9.33 * 10^{11}$

Correct answer:

0.170

Explanation:

Use the Henderson-Hasselbalch equation to determine the ratio of bicarbonate to carbonic acid in solution:

$pH = pK_a + \log\frac{\left [A^- \right ]}{\left [HA \right ]}$

$pH = pK_a + \log\frac{[HCO_3^-]}{[H_2CO_3]}$

$5.60-6.37 = -0.77 = \log\frac{\left [HCO_3^- \right ]}{\left [H_2CO_3 \right ]}$

Solve for the ratio we need to answer the question:

$10^{-0.77} = \frac{\left [ HCO_3^-\right ]}{[H_2CO_3]} = 0.170$

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AP Chemistry : Solutions

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #31 : Solutions

Example Question #32 : Solutions

Example Question #33 : Solutions

Example Question #34 : Solutions

Example Question #1 : Precipitates And Calculations

Example Question #1 : Precipitates And Calculations

Example Question #3 : Precipitates And Calculations

Example Question #4 : Precipitates And Calculations

Example Question #1 : Precipitates And Calculations

Example Question #6 : Precipitates And Calculations

All AP Chemistry Resources

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