College Chemistry Quiz: Representations Of Solutions

20 questions · exam conditions

0:00

Representations Of SolutionsQuestion 1 of 20

A solution contains $2.5 \times 10^{22}$ molecules of ethanol ( $\text{C}_2\text{H}_5\text{OH}$ ) dissolved in $500.0 \text{ mL}$ of solution. What is the molarity of this ethanol solution?

0.042 \text{ M}

0.083 \text{ M}

0.125 \text{ M}

0.208 \text{ M}

0.250 \text{ M}

← Back to quizzes

College Chemistry Quiz

College Chemistry Quiz: Representations Of Solutions

Practice Representations Of Solutions in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Representations Of Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A solution contains $2.5 \times 10^{22}$ molecules of ethanol ( $\text{C}_2\text{H}_5\text{OH}$ ) dissolved in $500.0 \text{ mL}$ of solution. What is the molarity of this ethanol solution?

$0.042 \text{ M}$
$0.083 \text{ M}$ (correct answer)
$0.125 \text{ M}$
$0.208 \text{ M}$
$0.250 \text{ M}$

Explanation: When you encounter problems asking for molarity from the number of molecules, you need to bridge the gap between the molecular scale and the molar scale using Avogadro's number. Molarity equals moles of solute divided by liters of solution. You have

2.5 \times 10^{22}

molecules of ethanol in

500.0 \text{ mL}

of solution. First, convert molecules to moles using Avogadro's number (

6.022 \times 10^{23}

molecules/mol):

\text{moles} = \frac{2.5 \times 10^{22} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} = 0.0415 \text{ mol}

Next, convert volume to liters:

500.0 \text{ mL} = 0.5000 \text{ L}

Finally, calculate molarity:

M = \frac{0.0415 \text{ mol}}{0.5000 \text{ L}} = 0.083 \text{ M}

This confirms answer B is correct. Answer A (

0.042 \text{ M}

) results from forgetting to convert mL to L—dividing moles by 500 instead of 0.5. Answer C (

0.125 \text{ M}

) likely comes from incorrectly using

4.0 \times 10^{23}

instead of Avogadro's number in the conversion. Answer D (

0.208 \text{ M}

) represents a double error: using the wrong Avogadro's number and failing to convert volume units. Study tip: Always write out the unit conversion steps explicitly. Molarity problems require three conversions: molecules → moles (using Avogadro's number), mL → L, then apply the molarity formula. Double-check that you're using

6.022 \times 10^{23}

and converting volume to liters.

Question 2

When $25.0 \text{ mL}$ of $0.400 \text{ M NaCl}$ is mixed with $75.0 \text{ mL}$ of $0.200 \text{ M NaCl}$ , what is the molarity of $\text{Cl}^-$ ions in the final solution?

$0.250 \text{ M}$ (correct answer)
$0.300 \text{ M}$
$0.350 \text{ M}$
$0.400 \text{ M}$
$0.600 \text{ M}$

Explanation: When you encounter dilution problems involving ionic compounds, you need to track both the total moles of ions and the final volume to find the new concentration. First, calculate the moles of

\text{Cl}^-

from each solution. Since NaCl dissociates completely into

\text{Na}^+

and

\text{Cl}^-

, each mole of NaCl produces one mole of chloride ions. From the first solution:

0.0250 \text{ L} \times 0.400 \text{ M} = 0.0100 \text{ mol Cl}^-

From the second solution:

0.0750 \text{ L} \times 0.200 \text{ M} = 0.0150 \text{ mol Cl}^-

Total moles of

\text{Cl}^-

0.0100 + 0.0150 = 0.0250 \text{ mol}

The final volume is

25.0 + 75.0 = 100.0 \text{ mL} = 0.1000 \text{ L}

Therefore:

[\text{Cl}^-] = \frac{0.0250 \text{ mol}}{0.1000 \text{ L}} = 0.250 \text{ M}

Answer A (

0.250 \text{ M}

) is correct. Answer B (

0.300 \text{ M}

) likely comes from incorrectly averaging the two concentrations:

\frac{0.400 + 0.200}{2} = 0.300

. This ignores the different volumes. Answer C (

0.350 \text{ M}

) might result from calculation errors in the weighted average approach. Answer D (

0.400 \text{ M}

) incorrectly assumes the concentration equals that of the more concentrated solution. Remember: in dilution problems, always calculate total moles first, then divide by the total final volume. Simple averaging only works when volumes are equal.

Question 3

A student needs to prepare $250.0 \text{ mL}$ of $0.125 \text{ M}$ sucrose solution from a stock solution of $0.500 \text{ M}$ sucrose. What volume of the stock solution should be used?

$31.3 \text{ mL}$
$62.5 \text{ mL}$ (correct answer)
$125 \text{ mL}$
$200 \text{ mL}$
$500 \text{ mL}$

Explanation: This is a dilution problem, one of the most common calculations you'll encounter in chemistry lab work. When you see a question asking for the volume of stock solution needed to make a more dilute solution, immediately think of the dilution formula:

M_1V_1 = M_2V_2

. Here, you're starting with a concentrated stock solution (

M_1 = 0.500 \text{ M}

) and need to find what volume of it (

V_1 = ?

) will give you your target solution (

M_2 = 0.125 \text{ M}

V_2 = 250.0 \text{ mL}

). Substituting into the formula:

(0.500 \text{ M})(V_1) = (0.125 \text{ M})(250.0 \text{ mL})

. Solving for

V_1

V_1 = \frac{(0.125)(250.0)}{0.500} = \frac{31.25}{0.500} = 62.5 \text{ mL}

. This confirms answer B. Looking at the wrong answers: A (31.3 mL) comes from incorrectly calculating

\frac{31.25}{1}

instead of dividing by 0.500, or from other arithmetic errors. C (125 mL) results from confusing the target molarity (0.125 M) with the volume needed, a common conceptual mistake. D (200 mL) might come from incorrectly using

\frac{250.0 \times 0.500}{0.125}

, essentially inverting the dilution relationship. Remember this pattern: you always need less volume of the concentrated solution than your final volume, and the more you're diluting (larger ratio between

M_1

and

M_2

), the smaller volume you'll need. Here you're diluting 4-fold (0.500/0.125 = 4), so you need ¼ of 250 mL.

Question 4

A stock solution of $\text{CuSO}_4$ has a concentration of $0.250 \text{ M}$ . If $50.0 \text{ mL}$ of this stock solution is diluted to a final volume of $500.0 \text{ mL}$ , what is the concentration of $\text{Cu}^{2+}$ ions in the diluted solution?

$0.0125 \text{ M}$
$0.0250 \text{ M}$ (correct answer)
$0.0500 \text{ M}$
$0.125 \text{ M}$
$0.250 \text{ M}$

Explanation: When you encounter dilution problems, you're working with the principle that the amount of solute (moles) remains constant—only the volume changes. The key relationship is

M_1V_1 = M_2V_2

, where the subscripts represent initial and final conditions. Starting with your stock solution:

M_1 = 0.250 \text{ M}

and

V_1 = 50.0 \text{ mL}

. After dilution:

V_2 = 500.0 \text{ mL}

and

M_2 = ?

Solving:

M_2 = \frac{M_1V_1}{V_2} = \frac{(0.250 \text{ M})(50.0 \text{ mL})}{500.0 \text{ mL}} = 0.0250 \text{ M}

Since

\text{CuSO}_4

dissociates completely in water (

\text{CuSO}_4 \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-}

), the concentration of

\text{Cu}^{2+}

ions equals the concentration of

\text{CuSO}_4

. Therefore,

[\text{Cu}^{2+}] = 0.0250 \text{ M}

, making B correct. Option A (

0.0125 \text{ M}

) represents half the correct value—you might get this if you incorrectly assumed only half the copper dissociates. Option C (

0.0500 \text{ M}

) doubles the correct answer, possibly from confusing which volume goes in the numerator versus denominator. Option D (

0.125 \text{ M}

) suggests using an incorrect dilution factor or mathematical error in the calculation. Remember: dilutions always decrease concentration, and the dilution factor here is 10-fold (

500 \div 50 = 10

), so your final concentration should be one-tenth of the original. Always check that your answer makes logical sense with the degree of dilution performed.

Question 5

Which representation best describes what happens when solid $\text{K}_2\text{SO}_4$ dissolves in water?

$\text{K}_2\text{SO}_4(s) \rightarrow \text{K}_2\text{SO}_4(aq)$ with molecules remaining intact in solution
$\text{K}_2\text{SO}_4(s) \rightarrow \text{K}^+(aq) + \text{SO}_4^{2-}(aq)$ producing equal moles of cations and anions
$\text{K}_2\text{SO}_4(s) \rightarrow 2\text{K}^+(aq) + \text{SO}_4^{2-}(aq)$ with complete dissociation into constituent ions (correct answer)
$\text{K}_2\text{SO}_4(s) \rightarrow \text{K}_2^{2+}(aq) + \text{SO}_4^{2-}(aq)$ maintaining the cation pairing in solution
$\text{K}_2\text{SO}_4(s) \rightarrow 2\text{K}^+(aq) + \text{S}^{6+}(aq) + 4\text{O}^{2-}(aq)$ with breakdown of all polyatomic ions

Explanation: When ionic compounds dissolve in water, you need to understand the process of dissociation—how the solid breaks apart into its constituent ions and what the correct stoichiometry looks like in the balanced equation. Potassium sulfate is an ionic compound composed of potassium ions (K⁺) and sulfate ions (SO₄²⁻). When it dissolves, water molecules surround and separate these ions completely. The key is getting the stoichiometry right: the formula K₂SO₄ tells you there are 2 potassium ions for every 1 sulfate ion, so the dissociation equation must reflect this ratio. Choice C correctly shows

\text{K}_2\text{SO}_4(s) \rightarrow 2\text{K}^+(aq) + \text{SO}_4^{2-}(aq)

, with the coefficient 2 in front of K⁺ matching the subscript in the formula and representing complete dissociation. Choice A is wrong because ionic compounds don't remain as intact molecules in aqueous solution—they dissociate into ions. Choice B has incorrect stoichiometry; it shows equal moles of cations and anions (1:1 ratio) when the actual ratio is 2:1. Choice D creates a fictional "K₂²⁺" ion, which doesn't exist—potassium only forms K⁺ ions, and the two potassium atoms don't stay paired together in solution. Remember this pattern: when writing dissociation equations for ionic compounds, the coefficients in your products must match the subscripts in the original formula. This ensures you're conserving both mass and charge while showing the true ionic nature of the dissolved compound.

Question 6

A solution is prepared by dissolving $0.50 \text{ mol}$ of glucose and $0.25 \text{ mol}$ of $\text{NaCl}$ in $1.0 \text{ kg}$ of water. What is the total molality of all dissolved particles?

$0.75 \text{ m}$
$1.0 \text{ m}$ (correct answer)
$1.25 \text{ m}$
$1.5 \text{ m}$
$2.0 \text{ m}$

Explanation: When you encounter molality problems involving ionic compounds, remember that molality measures moles of particles per kilogram of solvent, and ionic compounds dissociate into multiple particles. First, determine how many particles each solute produces. Glucose is a molecular compound that doesn't dissociate, so 0.50 mol of glucose produces 0.50 mol of particles. However, NaCl is an ionic compound that dissociates completely:

\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

. Therefore, 0.25 mol of NaCl produces 0.50 mol of particles (0.25 mol Na⁺ + 0.25 mol Cl⁻). The total moles of particles = 0.50 mol (glucose) + 0.50 mol (from NaCl) = 1.0 mol particles. Since molality = moles of particles ÷ kg of solvent, and you have 1.0 kg of water: molality = 1.0 mol ÷ 1.0 kg = 1.0 m. Choice A (0.75 m) incorrectly adds only the original moles of solutes without accounting for NaCl's dissociation. Choice C (1.25 m) might result from miscounting the particles from NaCl dissociation. Choice D (1.5 m) could come from incorrectly assuming NaCl produces three particles instead of two. The key study tip: Always identify whether compounds are molecular (no dissociation) or ionic (dissociation occurs), then count the actual particles produced. For simple salts like NaCl, remember the dissociation doubles the particle count.

Question 7

What volume of $0.125 \text{ M}$ $\text{CaCl}_2$ solution contains the same number of chloride ions as $75.0 \text{ mL}$ of $0.200 \text{ M}$ $\text{NaCl}$ solution?

$30.0 \text{ mL}$
$60.0 \text{ mL}$ (correct answer)
$75.0 \text{ mL}$
$120 \text{ mL}$
$150 \text{ mL}$

Explanation: When you encounter problems involving different compounds that produce the same ion, focus on the stoichiometry—how many ions each compound produces per formula unit. To find the volume of

\text{CaCl}_2

solution needed, first calculate the moles of chloride ions in the

\text{NaCl}

solution. Since

\text{NaCl}

produces one

\text{Cl}^-

ion per formula unit: Moles of

\text{Cl}^-

from

\text{NaCl}

0.0750 \text{ L} \times 0.200 \text{ M} \times 1 = 0.0150 \text{ mol Cl}^-

Now,

\text{CaCl}_2

produces two

\text{Cl}^-

ions per formula unit, so: Moles of

\text{CaCl}_2

needed =

\frac{0.0150 \text{ mol Cl}^-}{2} = 0.00750 \text{ mol CaCl}_2

Using

V = \frac{\text{moles}}{\text{molarity}}

V = \frac{0.00750 \text{ mol}}{0.125 \text{ M}} = 0.0600 \text{ L} = 60.0 \text{ mL}

The answer is B)

60.0 \text{ mL}

. Choice A (

30.0 \text{ mL}

) would result if you incorrectly divided by 4 instead of 2, perhaps confusing the stoichiometric relationships. Choice C (

75.0 \text{ mL}

) assumes equal volumes are needed, ignoring that

\text{CaCl}_2

produces twice as many chloride ions. Choice D (

120 \text{ mL}

) results from forgetting that

\text{CaCl}_2

produces two chloride ions and treating it as if it produces only one. Remember: always account for stoichiometry when comparing ionic solutions. The number of ions produced per formula unit directly affects the calculations, so identify this relationship before setting up your math.

Question 8

A student needs to prepare a solution containing $1.5 \times 10^{21}$ formula units of $\text{MgSO}_4$ in $200.0 \text{ mL}$ of solution. What molarity should be prepared?

$0.0062 \text{ M}$
$0.012 \text{ M}$ (correct answer)
$0.025 \text{ M}$
$0.050 \text{ M}$
$0.10 \text{ M}$

Explanation: This question tests your ability to convert between formula units and molarity, which requires understanding Avogadro's number and the molarity equation. When you see formula units in a concentration problem, you need to convert to moles first, then apply the molarity formula. Start by converting formula units to moles using Avogadro's number (

6.022 \times 10^{23}

formula units/mol):

\text{moles} = \frac{1.5 \times 10^{21} \text{ formula units}}{6.022 \times 10^{23} \text{ formula units/mol}} = 0.00249 \text{ mol}

Next, convert the volume to liters:

200.0 \text{ mL} = 0.2000 \text{ L}

Apply the molarity equation:

M = \frac{\text{moles}}{\text{liters}} = \frac{0.00249 \text{ mol}}{0.2000 \text{ L}} = 0.012 \text{ M}

This confirms answer B is correct. Looking at the wrong answers: A (

0.0062 \text{ M}

) results from forgetting to convert mL to L, giving you

\frac{0.00249}{0.2000} \div 100

. C (

0.025 \text{ M}

) comes from using an incorrect conversion factor or rounding error early in the calculation. D (

0.050 \text{ M}

) suggests a more significant calculation error, possibly using the wrong power of 10 in Avogadro's number. Study tip: Always write out your units during conversions—this catches mistakes like forgetting the mL to L conversion. Remember the sequence: formula units → moles (using Avogadro's number) → molarity (using volume in liters).

Question 9

A student mixes equal volumes of $0.20 \text{ M}$ $\text{NaOH}$ and $0.30 \text{ M}$ $\text{HCl}$ . Which statement best describes the final solution composition?

The solution contains equal concentrations of $\text{Na}^+$ and $\text{Cl}^-$ ions with neutral pH
The solution contains $0.10 \text{ M}$ $\text{Na}^+$ and $0.15 \text{ M}$ $\text{Cl}^-$ with excess $\text{H}^+$ ions (correct answer)
The solution contains $0.25 \text{ M}$ $\text{NaCl}$ with no excess acid or base present
The solution contains molecular $\text{NaOH}$ and $\text{HCl}$ in equilibrium with their dissociated forms
The solution contains only $\text{H}_2\text{O}$ molecules due to complete neutralization of all ionic species

Explanation: When you encounter acid-base neutralization problems, you need to determine which reactant is limiting and calculate the final concentrations after dilution. The reaction here is:

\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Since equal volumes are mixed, let's say 1 L of each. You start with 0.20 mol NaOH and 0.30 mol HCl. The stoichiometry is 1:1, so NaOH is the limiting reactant. After neutralization, you'll have consumed all 0.20 mol of NaOH, leaving 0.10 mol excess HCl, and producing 0.20 mol NaCl. The total volume is now 2 L, so the final concentrations are:

[\text{Na}^+] = 0.20 \text{ mol}/2 \text{ L} = 0.10 \text{ M}

[\text{Cl}^-] = 0.30 \text{ mol}/2 \text{ L} = 0.15 \text{ M}

, and

[\text{H}^+] = 0.10 \text{ mol}/2 \text{ L} = 0.05 \text{ M}

. This matches answer B perfectly. Answer A is wrong because the concentrations aren't equal, and the solution isn't neutral due to excess acid. Answer C incorrectly suggests 0.25 M NaCl (the actual concentration is 0.10 M) and misses the excess acid. Answer D is wrong because these are strong acids and bases that dissociate completely in aqueous solution—there's no equilibrium with molecular forms. The key strategy: always identify the limiting reactant first, then account for dilution when calculating final concentrations. Remember that mixing equal volumes means each original concentration gets halved, but the limiting reactant determines what's left over.

Question 10

An aqueous solution is prepared by dissolving $10.0 \text{ g}$ of $\text{CuCl}_2 \cdot 2\text{H}_2\text{O}$ in enough water to make $250.0 \text{ mL}$ of solution. What is the concentration of $\text{Cu}^{2+}$ ions?

$0.118 \text{ M}$
$0.236 \text{ M}$ (correct answer)
$0.295 \text{ M}$
$0.472 \text{ M}$
$0.590 \text{ M}$

Explanation: When you encounter molarity problems involving hydrated compounds, you need to carefully track what ions are actually produced when the compound dissolves in water. To find the

\text{Cu}^{2+}

concentration, start by calculating moles of

\text{CuCl}_2 \cdot 2\text{H}_2\text{O}

. The molar mass is: Cu (63.55) + 2Cl (70.90) + 4H (4.03) + 2O (32.00) = 170.48 g/mol. Moles =

\frac{10.0 \text{ g}}{170.48 \text{ g/mol}} = 0.0587 \text{ mol}

When

\text{CuCl}_2 \cdot 2\text{H}_2\text{O}

dissolves, it dissociates as:

\text{CuCl}_2 \cdot 2\text{H}_2\text{O} \rightarrow \text{Cu}^{2+} + 2\text{Cl}^- + 2\text{H}_2\text{O}

Each mole of the hydrated compound produces exactly one mole of

\text{Cu}^{2+}

ions, so you have 0.0587 mol of

\text{Cu}^{2+}

. Molarity =

\frac{0.0587 \text{ mol}}{0.250 \text{ L}} = 0.236 \text{ M}

Choice A (0.118 M) represents half the correct value—this would result from incorrectly thinking each formula unit produces 0.5 moles of

\text{Cu}^{2+}

. Choice C (0.295 M) comes from using an incorrect molar mass, likely forgetting the water molecules. Choice D (0.472 M) is double the correct answer, possibly from confusing the 1:1 ratio of compound to

\text{Cu}^{2+}

ions with the 1:2 ratio for chloride ions. The answer is B (0.236 M). Remember: hydrated compounds release the same number of metal ions as their anhydrous counterparts—the water molecules don't affect the ion ratios, just the molar mass.

Question 11

A solution contains $0.040 \text{ mol}$ of $\text{Na}_2\text{SO}_4$ dissolved in $500.0 \text{ g}$ of water. What is the molality of $\text{SO}_4^{2-}$ ions?

$0.040 \text{ m}$
$0.080 \text{ m}$ (correct answer)
$0.10 \text{ m}$
$0.16 \text{ m}$
$0.20 \text{ m}$

Explanation: When you encounter molality problems involving ionic compounds, remember that molality measures moles of solute particles per kilogram of solvent, and ionic compounds dissociate to produce multiple ions. To find the molality of

\text{SO}_4^{2-}

ions, you need to determine how many moles of sulfate ions are produced when

\text{Na}_2\text{SO}_4

dissolves. Sodium sulfate dissociates according to:

\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}

. This means each mole of

\text{Na}_2\text{SO}_4

produces exactly 1 mole of

\text{SO}_4^{2-}

ions. Starting with

0.040 \text{ mol}

\text{Na}_2\text{SO}_4

, you get

0.040 \text{ mol}

\text{SO}_4^{2-}

ions. The solvent mass is

500.0 \text{ g} = 0.5000 \text{ kg}

. Therefore: molality

= \frac{0.040 \text{ mol}}{0.5000 \text{ kg}} = 0.080 \text{ m}

Answer A (

0.040 \text{ m}

) incorrectly uses the original moles of

\text{Na}_2\text{SO}_4

without considering what specific ion the question asks for. Answer C (

0.10 \text{ m}

) likely results from using

400 \text{ g}

instead of

500 \text{ g}

as the solvent mass. Answer D (

0.16 \text{ m}

) incorrectly assumes that 1 mole of

\text{Na}_2\text{SO}_4

produces 2 moles of

\text{SO}_4^{2-}

ions, confusing the sulfate ion count with the sodium ion count. Always write out the dissociation equation first—this prevents confusion about how many of each ion type are produced per formula unit of the original compound.

Question 12

What volume of water must be added to $100.0 \text{ mL}$ of $0.800 \text{ M HCl}$ to prepare a $0.200 \text{ M HCl}$ solution?

$200.0 \text{ mL}$
$300.0 \text{ mL}$ (correct answer)
$400.0 \text{ mL}$
$500.0 \text{ mL}$
$600.0 \text{ mL}$

Explanation: This is a dilution problem where you're adding water to decrease the concentration of an acid solution. The key relationship to remember is that the moles of solute remain constant during dilution - only the total volume changes. Use the dilution equation:

M_1V_1 = M_2V_2

, where

M_1

and

V_1

are the initial molarity and volume, and

M_2

and

V_2

are the final molarity and volume. Plugging in your values:

(0.800 \text{ M})(100.0 \text{ mL}) = (0.200 \text{ M})(V_2)

. Solving for

V_2

V_2 = \frac{80.0}{0.200} = 400.0 \text{ mL}

. This is the final total volume of the diluted solution. Since you started with

100.0 \text{ mL}

, the volume of water added is

400.0 - 100.0 = 300.0 \text{ mL}

. Choice A (

200.0 \text{ mL}

) would give you a final volume of only

300.0 \text{ mL}

, resulting in a concentration that's too high. Choice C (

400.0 \text{ mL}

) is the trap answer - this is the final total volume, not the volume of water added. Choice D (

500.0 \text{ mL}

) would create a final volume of

600.0 \text{ mL}

, making the solution too dilute. The correct answer is B. Study tip: Always distinguish between "final volume" and "volume added" in dilution problems. The dilution equation gives you the final volume, so don't forget to subtract the initial volume to find how much solvent you need to add.

Question 13

A solution is prepared by dissolving $15.0 \text{ g}$ of $\text{KBr}$ in $85.0 \text{ g}$ of water. What is the mass percent of $\text{KBr}$ in this solution?

$15.0\%$ (correct answer)
$17.6\%$
$25.0\%$
$85.0\%$
$150\%$

Explanation: When you encounter mass percent problems, you're dealing with concentration calculations that require careful attention to what constitutes the total mass of the solution. Mass percent is defined as:

\text{Mass percent} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100\%

Here, you have 15.0 g of KBr (solute) dissolved in 85.0 g of water (solvent). The crucial step is finding the total mass of the solution, which is the sum of solute and solvent masses: 15.0 g + 85.0 g = 100.0 g. Now you can calculate:

\text{Mass percent of KBr} = \frac{15.0 \text{ g}}{100.0 \text{ g}} \times 100\% = 15.0\%

This confirms that answer A (15.0%) is correct. Answer B (17.6%) represents a common error where students divide the solute mass by only the solvent mass:

\frac{15.0}{85.0} \times 100\% = 17.6\%

. This gives mass of solute per mass of solvent, not mass percent of the solution. Answer C (25.0%) might result from incorrectly using ratios like

\frac{15.0}{60.0}

, possibly from arithmetic errors in finding the total mass. Answer D (85.0%) would be the mass percent of water, not KBr, representing a mix-up of solute and solvent. Remember: mass percent always uses the total solution mass as the denominator, never just the solvent mass. The percentages of all components in a solution must add up to 100%.

Question 14

A student prepares a $0.15 \text{ M}$ aqueous solution of glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) by dissolving $27.0 \text{ g}$ of glucose in water. What is the total volume of the solution prepared?

$0.833 \text{ L}$
$1.00 \text{ L}$ (correct answer)
$1.20 \text{ L}$
$1.50 \text{ L}$
$2.70 \text{ L}$

Explanation: This question tests your understanding of molarity calculations, which relate the amount of solute (in moles) to the total volume of solution. When you see molarity problems, remember the formula:

M = \frac{\text{moles of solute}}{\text{liters of solution}}

To find the solution volume, you need to first convert the mass of glucose to moles, then use the molarity formula. The molar mass of glucose (

\text{C}_6\text{H}_{12}\text{O}_6

) is

6(12.01) + 12(1.008) + 6(16.00) = 180.16 \text{ g/mol}

. Converting 27.0 g to moles:

\frac{27.0 \text{ g}}{180.16 \text{ g/mol}} = 0.150 \text{ mol}

. Now using the molarity equation:

0.15 \text{ M} = \frac{0.150 \text{ mol}}{V}

, so

V = \frac{0.150 \text{ mol}}{0.15 \text{ M}} = 1.00 \text{ L}

. This confirms answer B. Answer A (0.833 L) results from incorrectly using 27.0 g directly without converting to moles. Answer C (1.20 L) likely comes from calculation errors or using an incorrect molar mass. Answer D (1.50 L) might result from confusing the given molarity (0.15 M) with the calculated moles (0.150 mol) in the denominator. Always work systematically through molarity problems: convert mass to moles using molar mass, then apply the molarity formula. Double-check your molar mass calculation, as this is a common source of error in these problems.

Question 15

A solution is prepared by mixing $100.0 \text{ mL}$ of $0.150 \text{ M}$ $\text{MgCl}_2$ with $200.0 \text{ mL}$ of $0.100 \text{ M}$ $\text{MgCl}_2$ . What is the final molarity of $\text{Cl}^-$ ions?

$0.100 \text{ M}$
$0.117 \text{ M}$
$0.200 \text{ M}$
$0.233 \text{ M}$ (correct answer)
$0.467 \text{ M}$

Explanation: When you encounter solution mixing problems involving ionic compounds, you need to track both the dilution effect and the stoichiometry of ion formation. This question tests your ability to calculate final ion concentrations after mixing solutions of different concentrations. First, calculate the moles of

\text{MgCl}_2

from each solution. The first solution contributes

(0.1000 \text{ L})(0.150 \text{ M}) = 0.0150 \text{ mol}

\text{MgCl}_2

. The second solution contributes

(0.2000 \text{ L})(0.100 \text{ M}) = 0.0200 \text{ mol}

\text{MgCl}_2

. Total moles of

\text{MgCl}_2 = 0.0350 \text{ mol}

. Since each

\text{MgCl}_2

molecule produces 2 chloride ions when it dissociates, the total moles of

\text{Cl}^-

ions equals

0.0350 \text{ mol} \times 2 = 0.0700 \text{ mol}

. The final volume is

100.0 + 200.0 = 300.0 \text{ mL} = 0.3000 \text{ L}

. Therefore, the molarity of

\text{Cl}^-

\frac{0.0700 \text{ mol}}{0.3000 \text{ L}} = 0.233 \text{ M}

. Choice A (

0.100 \text{ M}

) represents the concentration of the more dilute

\text{MgCl}_2

solution, ignoring both the 2:1 stoichiometry and the mixing effect. Choice B (

0.117 \text{ M}

) correctly calculates the final

\text{MgCl}_2

concentration but forgets to account for the two chloride ions per formula unit. Choice C (

0.200 \text{ M}

) might result from calculation errors in the mole or volume calculations. Remember: always account for ion stoichiometry in dissociation reactions. Polyatomic salts like

\text{MgCl}_2

don't produce a 1:1 ratio of cation to anion.

Question 16

In which of the following aqueous solutions would the solute particles be best represented as individual molecules rather than ions?

$\text{NaCl}$ solution showing separate $\text{Na}^+$ and $\text{Cl}^-$ ions surrounded by water molecules
$\text{HCl}$ solution showing $\text{H}_3\text{O}^+$ and $\text{Cl}^-$ ions with hydrogen bonding to water
$\text{C}_6\text{H}_{12}\text{O}_6$ solution showing intact glucose molecules hydrogen bonding with water (correct answer)
$\text{KOH}$ solution showing separated $\text{K}^+$ and $\text{OH}^-$ ions in aqueous environment
$\text{CaCl}_2$ solution showing one $\text{Ca}^{2+}$ ion and two $\text{Cl}^-$ ions per formula unit dissolved

Explanation: When you encounter questions about how substances behave in aqueous solutions, you need to distinguish between compounds that ionize (break apart into charged particles) versus those that remain as intact molecules when dissolved in water. The key is understanding that ionic compounds and strong acids/bases dissociate completely in water, while molecular compounds like sugars typically dissolve without breaking apart. Choice C shows glucose (

\text{C}_6\text{H}_{12}\text{O}_6

) remaining as complete molecules that form hydrogen bonds with water. Glucose is a covalent compound that dissolves by interacting with water molecules through hydrogen bonding, but the glucose molecules stay intact—they don't break into smaller charged pieces. Let's examine why the other options show ionized particles: Choice A depicts

\text{NaCl}

, an ionic compound that completely dissociates into

\text{Na}^+

and

\text{Cl}^-

ions when dissolved. Choice B shows

\text{HCl}

, a strong acid that donates protons to water, forming

\text{H}_3\text{O}^+

and

\text{Cl}^-

ions. Choice D represents

\text{KOH}

, a strong base that completely ionizes into

\text{K}^+

and

\text{OH}^-

ions. Remember this pattern: ionic compounds (metal + nonmetal), strong acids, and strong bases will exist as separate ions in aqueous solution, while molecular compounds like sugars, alcohols, and weak electrolytes often remain as intact molecules. When you see molecular formulas for organic compounds like glucose, think "stays together" rather than "breaks apart."

Question 17

Which of the following correctly represents the dissolution of $\text{Al}(\text{NO}_3)_3$ in water?

$\text{Al}(\text{NO}_3)_3(s) \rightarrow \text{Al}^{3+}(aq) + \text{NO}_3^-(aq)$ producing equal molar amounts of cation and anion
$\text{Al}(\text{NO}_3)_3(s) \rightarrow \text{Al}^{3+}(aq) + 3\text{NO}_3^-(aq)$ with complete dissociation into constituent ions (correct answer)
$\text{Al}(\text{NO}_3)_3(s) \rightarrow 3\text{Al}^{3+}(aq) + \text{NO}_3^-(aq)$ balancing the total positive and negative charges
$\text{Al}(\text{NO}_3)_3(s) \rightarrow \text{Al}^{3+}(aq) + 3\text{N}^{5+}(aq) + 9\text{O}^{2-}(aq)$ with complete breakdown of polyatomic ions
$\text{Al}(\text{NO}_3)_3(s) \rightarrow \text{Al}(\text{NO}_3)_3(aq)$ remaining as intact molecules in aqueous solution

Explanation: When ionic compounds dissolve in water, they dissociate into their constituent ions. The key is identifying what ions are actually present in the original compound and writing a balanced equation that conserves both mass and charge. Let's analyze

\text{Al}(\text{NO}_3)_3

: it contains one aluminum ion (

\text{Al}^{3+}

) and three nitrate ions (

\text{NO}_3^-

). When it dissolves, these ions separate completely, giving us:

\text{Al}(\text{NO}_3)_3(s) \rightarrow \text{Al}^{3+}(aq) + 3\text{NO}_3^-(aq)

. This equation is balanced because the total charge on both sides is zero: one

\text{Al}^{3+}

(+3 charge) plus three

\text{NO}_3^-

(-3 total charge). Choice A incorrectly shows only one nitrate ion instead of three. The subscript 3 in the formula tells you there are three

\text{NO}_3^-

ions per formula unit, not one. Choice C incorrectly produces three aluminum ions instead of one - the aluminum doesn't have a subscript, so there's only one

\text{Al}^{3+}

per formula unit. Choice D makes the error of breaking down the polyatomic nitrate ion further into individual atoms. Nitrate (

\text{NO}_3^-

) is a stable polyatomic ion that stays intact when the compound dissolves - it doesn't break apart into separate nitrogen and oxygen atoms. Remember: when writing dissolution equations, polyatomic ions like nitrate, sulfate, and carbonate remain as complete units. Only break ionic compounds at the boundaries between cations and anions, never within polyatomic ions themselves.

Question 18

A solution contains $8.5 \text{ g}$ of $\text{NH}_3$ dissolved in $1.2 \text{ kg}$ of water. What is the molality of this solution?

$0.21 \text{ m}$
$0.42 \text{ m}$ (correct answer)
$0.50 \text{ m}$
$0.71 \text{ m}$
$7.1 \text{ m}$

Explanation: Molality questions test your understanding of concentration units that depend on the mass of solvent rather than solution volume. When you see a molality problem, remember the formula: molality = moles of solute ÷ kilograms of solvent. To solve this, you need to find the moles of

\text{NH}_3

first. The molar mass of ammonia is

14.0 + 3(1.0) = 17.0 \text{ g/mol}

. Converting the given mass:

\frac{8.5 \text{ g}}{17.0 \text{ g/mol}} = 0.50 \text{ mol}

\text{NH}_3

. Next, identify the solvent mass:

1.2 \text{ kg}

of water (already in the correct units). Now calculate molality:

\frac{0.50 \text{ mol}}{1.2 \text{ kg}} = 0.42 \text{ m}

, confirming answer B. Let's examine why the other options are incorrect. Choice A (

0.21 \text{ m}

) represents exactly half the correct answer, suggesting an error like using

2.4 \text{ kg}

instead of

1.2 \text{ kg}

for the solvent mass. Choice C (

0.50 \text{ m}

) equals the number of moles calculated, indicating the student forgot to divide by the solvent mass entirely. Choice D (

0.71 \text{ m}

) approximates what you'd get if you incorrectly used

0.7 \text{ kg}

as the solvent mass, possibly misreading the given

1.2 \text{ kg}

. The key to molality problems is organization: calculate moles of solute, confirm solvent mass is in kilograms, then divide. Don't confuse molality with molarity—molality uses solvent mass, not solution volume.

Question 19

A student dissolves $5.85 \text{ g}$ of $\text{NaCl}$ in enough water to make $250.0 \text{ mL}$ of solution. What is the concentration of $\text{Na}^+$ ions in this solution?

$0.200 \text{ M}$
$0.300 \text{ M}$
$0.400 \text{ M}$ (correct answer)
$0.500 \text{ M}$
$0.800 \text{ M}$

Explanation: When you encounter molarity problems involving ionic compounds, remember that molarity measures moles of solute per liter of solution, and you must account for how many ions each formula unit produces. To find the

\text{Na}^+

concentration, start by calculating the molarity of

\text{NaCl}

. First, convert grams to moles:

5.85 \text{ g NaCl} \times \frac{1 \text{ mol}}{58.5 \text{ g}} = 0.100 \text{ mol NaCl}

. Then calculate molarity:

\frac{0.100 \text{ mol}}{0.250 \text{ L}} = 0.400 \text{ M NaCl}

. Here's the crucial step:

\text{NaCl}

dissociates completely in water according to

\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

. Each formula unit produces exactly one

\text{Na}^+

ion, so the concentration of

\text{Na}^+

equals the molarity of

\text{NaCl}

0.400 \text{ M}

. Answer choice A (

0.200 \text{ M}

) represents half the correct molarity—you might get this if you incorrectly divided by 2, perhaps confusing this with a divalent ion situation. Choice B (

0.300 \text{ M}

) doesn't correspond to any logical calculation error. Choice D (

0.500 \text{ M}

) suggests using an incorrect molar mass, possibly

46.8 \text{ g/mol}

instead of

58.5 \text{ g/mol}

. The correct answer is C (

0.400 \text{ M}

). Study tip: Always write the dissociation equation for ionic compounds. For

\text{CaCl}_2

, you'd get 2 chloride ions per formula unit, doubling the chloride concentration compared to the compound's molarity. The stoichiometric coefficients in the dissociation equation are your multipliers.

Question 20

The graph shows the solubility of $\text{KNO}_3$ in water as a function of temperature. At $60°\text{C}$ , what mass of $\text{KNO}_3$ can be dissolved in $250 \text{ g}$ of water to form a saturated solution?

$55 \text{ g}$
$110 \text{ g}$
$220 \text{ g}$
$275 \text{ g}$ (correct answer)
$330 \text{ g}$

Explanation: From the graph, the solubility of

\text{KNO}_3

60°\text{C}

110 \text{ g}

per

100 \text{ g}

of water. For

250 \text{ g}

of water:

\frac{110 \text{ g KNO}_3}{100 \text{ g H}_2\text{O}} \times 250 \text{ g H}_2\text{O} = 275 \text{ g KNO}_3

. Choice A uses the solubility value for

100 \text{ g}

water divided by 2. Choice B uses the solubility value directly without scaling. Choice C uses double the solubility value. Choice E uses triple the solubility value without proper calculation.