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Example Questions

Example Question #411 : Ap Chemistry

What volume of water must be added to 750mL of 0.050M sodium chloride ( NaCl ) in order to achieve a final concentration of 0.015M?

Possible Answers:

2.5L

2.25L

1.75L

Correct answer:

1.75L

Explanation:

For a solution of known volume and concentration (molarity in this case), the volume needed to dilute the solution to a desired concentration may be found using the formula:

$M_{1}V_{1} = M_{2} V_2$

Where M_1 and M_2 are the initial and final concentrations, and V_1 and V_2 are the initial and final volumes. So, for 750mL (0.750L) of a 0.050M solution diluted to 0.015M:

0.050 M * 0.750 L = 0.015 M * V_2

Solving for V_2 :

$V_2 = \frac{0.050 M * 0.750L}{0.015 M} = 2.5 L$

Now that we know the total volume needed, we may find the volume that must be added by subtracting the initial volume ( V_1 ) from the final volume ( V_2 ):

V_2 - V_1 = 2.5 L - 0.750 L = 1.75 L

1.75L of water must be added to 750mL of 0.050M NaCl in order to achieve a final concentration of 0.015M

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Example Question #51 : Solutions

What is the pH of a 0.025M solution of hydrochloric acid ( HCl )?

Possible Answers:

1.06

-1.60

0.944

1.60

Correct answer:

1.60

Explanation:

Since HCl is a strong acid, calculations should be carried out assuming that the compound dissociates completely:

$HCl\rightarrow H^{+} + Cl^{-}$

H^+ and Cl^- are produced in a 1:1 ratio to total dissolved HCl , so the concentration of HCl in solution is the same as the concentration of H^+ :

$\left [ HCl\right ] = \left [ H^{+}\right ] = \left [ Cl^{-}\right ]$

pH is related to the concentration of H^+ :

$pH = -log[H^{+}]$

pH = -log[0.025] = 1.60

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Example Question #97 : Solutions And States Of Matter

What is the osmotic pressure of a 5.0M solution of $\small C_{2}H_{6}$ at $\small 10^{o}C$ ?

Possible Answers:

$\small 232 atm$

$\small 116 atm$

$\small 4 atm$

$\small 1 atm$

Correct answer:

$\small 116 atm$

Explanation:

Osmotic pressure is represented by:

$\small \Pi = iMRT$

Where $\small i=$ Van’t hoff factor, $\small M = Molarity$ , $\small R =$ gas constant $\left(0.08206 \frac{L\cdot atm}{mol\cdot K}\right)$ , $\small T =$ temperature in $\small K$ . The Van’t hoff factor is a unitless number that represents the amount of ionic species that the compound $\small (C_{2}H_{6})$ will dissociate in solution. $\small C_{2}H_{6}$ is part of a large group of molecules classified as hydrocarbons which normally do not dissociate at all in solution. Therefore, $\small i = 1$ .

Plug in known values and solve.

$\Pi = 1\cdot 5\cdot 0.08206\frac{L\cdot atm}{mol\cdot K}\cdot 283$

$\Pi= 116atm$

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Example Question #98 : Solutions And States Of Matter

A solution was prepared by dissolving 22.0 grams of NaOH in water to give a 110mL solution. What is the concentration in molarity of this solution?

Possible Answers:

0.55M

10M

25M

Correct answer:

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 NaOH to calculate the moles of solute:

$MW_{NaOH}=(23\frac{g}{mole})+(16\frac{g}{mole})+(1\frac{g}{mole})=40\frac{g}{mole}$

$22g*\frac{mole}{40g}=0.55\ moles\ NaOH$

$\frac{0.55\ moles\ NaOH}{0.110L}=5M$

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Example Question #411 : Ap Chemistry

A solution was prepared by dissolving 40.0 grams of $CH_{3}CH_{2}OH$ in water to give a 50mL solution. What is the concentration in molarity of this solution?

Possible Answers:

1.3M

17M

11M

10M

40M

Correct answer:

17M

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 $CH_{3}CH_{2}OH$ to calculate the moles of solute:

$MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ O)$

$MW_{CH_{3}CH_{2}OH}=2(12\frac{g}{mole})+6(1g\frac{g}{mole})+(16\frac{g}{mole})=46 \frac{g}{mole}$

$40.0g*\frac{mole}{46g}=0.87\ moles\ CH_{3}CH_{2}OH$

$\frac{0.87\ moles\ CH_{3}CH_{2}OH}{0.050\L}=17M$

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Example Question #91 : Solutions And States Of Matter

How many milliliters of solution is needed to dissolve 5 grams of NaCl to prepare a solution of concentration 10M?

Possible Answers:

15mL

8.6mL

42mL

20mL

1.5mL

Correct answer:

8.6mL

Explanation:

In order to calculate the number of milliliters, we must first determine the number of moles in 5 grams of NaCl using its molecular weight as a conversion factor:

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

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

$5g*\frac{mole}{58g}= 0.086\ moles$

Using the concentration units as a conversion factor and the number of moles calculated, the number of milliliters can be calculated:

$0.086 moles * \frac{L}{10\ moles}= 0.0086\ L = 8.6\ mL$

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Example Question #11 : Other Solution Concepts

A solution was prepared by diluting 10mL of a 0.500M salt solution to 20mL. What would be the final concentration of this solution?

Possible Answers:

0.10M

0.02M

0.25M

0.12M

0.45M

Correct answer:

0.25M

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:

$M_{2}=\frac{0.500M * 10mL}{20mL}= 0.25M$

Therefore, after diluting the solution to 20mL, the solution concentration would be lowered from 0.50M to 0.25M.

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Example Question #51 : Solutions

Which of the following is a weak electrolyte?

Possible Answers:

AgBr

NH_4Cl

H_2O_2

HNO_3

Correct answer:

H_2O_2

Explanation:

Solutes that dissociate completely in a solution are called strong electrolytes. Weak electrolytes stay paired to some extent in solutions. As a result, strong electrolytes include ionic compounds and strong acid and bases.

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Example Question #12 : Other Solution Concepts

Which of the following definitions is false?

Possible Answers:

Molality is the number of moles of solute in a solution divided by the number of kilogram of solvent.

Ion-product constant of water, $K_{w}$ , is the product of equilibrium concentration of H_3O^+ and OH^- ions in an aqueous solution at $25^\circ C$ .

Solubility product, $K_{sp}$ , is the product of ion concentrations at equilibrium in a supersaturated solution of salt.

The van't Hoff factor, i, is the number of ions that a compound produces in a solution.

Correct answer:

Solubility product, $K_{sp}$ , is the product of ion concentrations at equilibrium in a supersaturated solution of salt.

Explanation:

Solubility product, $K_{sp}$ , is the product of ion concentrations at equilibrium in a saturated solution of salt. All other definitions are true.

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Example Question #1 : Stoichiometry

Balance the following equation:

NaOH + H₂SO₄ → H₂O + Na₂SO₄

Possible Answers:

4 NaOH + 2 H₂SO₄ → 4 H₂O + 2 Na₂SO₄

2 NaOH + H₂SO₄ → H₂O + Na₂SO₄

NaOH + H₂SO₄ → H₂O + Na₂SO₄

2 NaOH + H₂SO₄ → 2 H₂O + Na₂SO₄

4 NaOH + H₂SO₄ → 2 H₂O + 2 Na₂SO₄

Correct answer:

2 NaOH + H₂SO₄ → 2 H₂O + Na₂SO₄

Explanation:

Balancing equations:

there is 1 NaOH and 1 H₂SO₄ in the reactants; and 1 H₂O and 1 Na₂SO₄ in the products

steps:

1) balance the Na first; get 2 NaOH

2) balance H next; get 1 H₂SO4, 2 H₂O

3) check to make sure that O and S balance

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AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #411 : Ap Chemistry

Example Question #51 : Solutions

Example Question #97 : Solutions And States Of Matter

Example Question #98 : Solutions And States Of Matter

Example Question #411 : Ap Chemistry

Example Question #91 : Solutions And States Of Matter

Example Question #11 : Other Solution Concepts

Example Question #51 : Solutions

Example Question #12 : Other Solution Concepts

Example Question #1 : Stoichiometry

All AP Chemistry Resources

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