College Chemistry Quiz: Solutions And Mixtures

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Solutions And MixturesQuestion 1 of 20

When 50.0 mL of 0.200 M $NaCl$ is mixed with 150.0 mL of 0.100 M $NaCl$ , what is the molarity of $NaCl$ in the final solution? Assume volumes are additive.

0.075 M

0.125 M

0.150 M

0.175 M

0.300 M

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College Chemistry Quiz

College Chemistry Quiz: Solutions And Mixtures

Practice Solutions And Mixtures 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 Solutions And Mixtures, 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

When 50.0 mL of 0.200 M $NaCl$ is mixed with 150.0 mL of 0.100 M $NaCl$ , what is the molarity of $NaCl$ in the final solution? Assume volumes are additive.

0.075 M
0.125 M (correct answer)
0.150 M
0.175 M
0.300 M

Explanation: This question tests your understanding of dilution and mixing solutions of different concentrations. When you mix two solutions containing the same solute, you need to account for the total moles of solute and the total volume to find the final molarity. Start by calculating the moles of

NaCl

in each solution. For the first solution:

0.0500 \text{ L} \times 0.200 \text{ M} = 0.0100 \text{ mol NaCl}

. For the second solution:

0.1500 \text{ L} \times 0.100 \text{ M} = 0.0150 \text{ mol NaCl}

. The total moles of

NaCl

0.0100 + 0.0150 = 0.0250 \text{ mol}

. The total volume is

50.0 + 150.0 = 200.0 \text{ mL} = 0.2000 \text{ L}

. Therefore, the final molarity is

\frac{0.0250 \text{ mol}}{0.2000 \text{ L}} = 0.125 \text{ M}

, which is answer choice B. Answer choice A (0.075 M) likely results from incorrectly averaging the two concentrations without considering the different volumes. Choice C (0.150 M) is simply the average of 0.200 M and 0.100 M, ignoring that there are different amounts of each solution. Choice D (0.175 M) might come from weighted averaging errors or calculation mistakes. Remember the key formula for mixing problems:

M_{\text{final}} = \frac{n_1 + n_2}{V_{\text{total}}}

. Always calculate total moles first, then divide by total volume. Don't just average concentrations unless the volumes are equal.

Question 2

A solution contains 15.0% by mass $KBr$ in water. If the density of this solution is 1.12 g/mL, what is the molarity of $KBr$ ? (Molar mass of $KBr$ = 119.0 g/mol)

1.26 M
1.41 M (correct answer)
1.68 M
2.52 M
9.41 M

Explanation: When you encounter percentage composition and density together, you're being asked to convert between different concentration units. This requires a systematic approach using the relationship between mass percentage, density, and molarity. To find molarity, you need moles of solute per liter of solution. Start by assuming 1 L of solution. Since density is 1.12 g/mL, this liter has a mass of 1120 g. With 15.0%

KBr

by mass, the solution contains

1120 \text{ g} \times 0.150 = 168 \text{ g}

KBr

. Converting to moles:

\frac{168 \text{ g}}{119.0 \text{ g/mol}} = 1.41 \text{ mol}

. Therefore, molarity =

\frac{1.41 \text{ mol}}{1 \text{ L}} = 1.41 \text{ M}

. Answer B (1.41 M) is correct based on this calculation. Answer A (1.26 M) likely results from calculation errors or using incorrect mass values. Answer C (1.68 M) could come from forgetting to account for the total solution mass or mixing up the percentage calculation. Answer D (2.52 M) is roughly double the correct answer, suggesting someone might have made an error in unit conversion or used the wrong denominator in their molarity calculation. Study tip: Always work systematically with mass percentage problems: assume a convenient volume (like 1 L), use density to find total mass, apply the percentage to get solute mass, convert to moles, then calculate molarity. Double-check that you're using the solution volume, not solvent volume, in your final calculation.

Question 3

A solution is prepared by mixing 200.0 mL of 0.300 M $CaCl_2$ with 300.0 mL of 0.800 M $NaCl$ . What is the molarity of $Cl^-$ ions in the final solution? Assume volumes are additive.

0.240 M
0.480 M
0.560 M
0.720 M (correct answer)
1.400 M

Explanation: This question tests your understanding of solution mixing and ion concentration calculations. When mixing ionic solutions, you need to track both the ions contributed by each compound and the change in total volume. Start by calculating moles of chloride ions from each source.

CaCl_2

dissociates into one

Ca^{2+}

and two

Cl^-

ions, so each mole of

CaCl_2

contributes 2 moles of

Cl^-

. From the

CaCl_2

solution: (0.200 L)(0.300 M) = 0.0600 mol

CaCl_2

, which gives 0.120 mol

Cl^-

. From the

NaCl

solution: (0.300 L)(0.800 M) = 0.240 mol

NaCl

, which gives 0.240 mol

Cl^-

(since

NaCl

produces one

Cl^-

per formula unit). Total

Cl^-

ions = 0.120 + 0.240 = 0.360 mol. The final volume is 200.0 + 300.0 = 500.0 mL = 0.500 L. Therefore, molarity of

Cl^-

= 0.360 mol ÷ 0.500 L = 0.720 M, confirming answer D. Answer A (0.240 M) likely comes from only counting

Cl^-

from

NaCl

and ignoring

CaCl_2

. Answer B (0.480 M) results from forgetting that

CaCl_2

produces two

Cl^-

ions, treating it like it produces only one. Answer C (0.560 M) appears to involve calculation errors in the mole or volume determinations. Remember: always account for the stoichiometry of ion production when compounds dissociate, and don't forget that mixing solutions changes the total volume for your final molarity calculation.

Question 4

A stock solution of $H_2SO_4$ has a concentration of 18.0 M. What volume of this stock solution is needed to prepare 500.0 mL of 0.500 M $H_2SO_4$ ?

13.9 mL (correct answer)
16.7 mL
25.0 mL
27.8 mL
36.0 mL

Explanation: This is a dilution problem, one of the most common types of solution preparation questions in chemistry. When you see a question asking about preparing a more dilute solution from a concentrated stock solution, you need the dilution equation:

M_1V_1 = M_2V_2

, where the subscripts 1 and 2 represent the initial (stock) and final (diluted) solutions. Given:

M_1 = 18.0 \text{ M}

M_2 = 0.500 \text{ M}

V_2 = 500.0 \text{ mL}

, and you need to find

V_1

. Rearranging the equation:

V_1 = \frac{M_2V_2}{M_1} = \frac{(0.500 \text{ M})(500.0 \text{ mL})}{18.0 \text{ M}} = \frac{250.0}{18.0} = 13.9 \text{ mL}

This confirms answer A is correct. Looking at the wrong answers: B (16.7 mL) might result from calculation errors or using incorrect values. C (25.0 mL) appears to come from dividing 500.0 by 20 instead of 18, suggesting confusion with the molarity values. D (27.8 mL) is exactly double the correct answer, indicating a possible factor-of-two error in the calculation. Remember this key strategy: always check that your answer makes sense conceptually. Since you're diluting from 18.0 M to 0.500 M (about a 36-fold dilution), you should need a relatively small volume of stock solution compared to your final volume. The correct answer of 13.9 mL from 500.0 mL total fits this expectation perfectly.

Question 5

What is the mole fraction of ethanol ( $C_2H_5OH$ ) in a solution prepared by mixing 46.0 g of ethanol (molar mass = 46.1 g/mol) with 54.0 g of water (molar mass = 18.0 g/mol)?

0.251 (correct answer)
0.333
0.460
0.540
0.749

Explanation: When you encounter mole fraction problems, you're dealing with a way to express concentration based on the ratio of moles of one component to the total moles in solution. The key is converting masses to moles, then finding the fraction. First, convert each component to moles. For ethanol:

\frac{46.0 \text{ g}}{46.1 \text{ g/mol}} = 0.998 \text{ mol}

. For water:

\frac{54.0 \text{ g}}{18.0 \text{ g/mol}} = 3.00 \text{ mol}

. The total moles is

0.998 + 3.00 = 3.998 \text{ mol}

. The mole fraction of ethanol is:

\chi_{ethanol} = \frac{0.998}{3.998} = 0.250

, which rounds to 0.251 (Answer A). Answer B (0.333) represents a common error where students incorrectly assume equal molar amounts or confuse this with a different concentration unit. Answer C (0.460) likely comes from using mass percentages instead of mole fractions—ethanol's mass percentage is

\frac{46.0}{100.0} = 0.460

. Answer D (0.540) similarly represents water's mass percentage, suggesting confusion between mass-based and mole-based concentration units. Remember that mole fraction problems always require three steps: convert masses to moles, find total moles, then divide the component's moles by the total. Watch out for the trap of using mass percentages instead—mole fractions specifically require molar quantities, not mass quantities.

Question 6

What volume of 0.500 M $Na_3PO_4$ solution contains 0.750 mol of $PO_4^{3-}$ ions?

0.375 L
0.750 L
1.50 L (correct answer)
2.25 L
3.75 L

Explanation: When you encounter molarity problems involving polyatomic ions, you need to carefully track the relationship between the compound's concentration and the concentration of its constituent ions. To find the volume, start with the molarity equation:

M = \frac{\text{moles of solute}}{\text{volume in L}}

. The key insight is identifying what your "solute" is. Since

Na_3PO_4

dissociates completely in water to produce

3Na^+ + PO_4^{3-}

, each formula unit of

Na_3PO_4

produces exactly one

PO_4^{3-}

ion. This means the molarity of

PO_4^{3-}

ions equals the molarity of the

Na_3PO_4

solution: 0.500 M. Now solve for volume:

0.500 \text{ M} = \frac{0.750 \text{ mol}}{V}

, so

V = \frac{0.750}{0.500} = 1.50 \text{ L}

. The answer is C. Looking at the wrong answers: A (0.375 L) results from incorrectly multiplying 0.750 mol by 0.500 M instead of dividing. B (0.750 L) comes from assuming you need 0.750 L to get 0.750 mol, ignoring the concentration entirely. D (2.25 L) suggests confusion about the stoichiometry—perhaps thinking you need three times as much volume because sodium phosphate contains three sodium ions. Remember this pattern: when calculating volumes from molarity, always identify which species you're tracking and use

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

. Pay special attention to the stoichiometric relationship between the compound and the ion of interest.

Question 7

What is the percent by mass of $NaOH$ in a solution prepared by dissolving 12.0 g of $NaOH$ in 88.0 g of water?

10.9%
12.0% (correct answer)
13.6%
15.0%
88.0%

Explanation: When you encounter percent by mass problems, you're calculating what fraction of the total solution mass comes from a specific component. This is a fundamental concentration calculation in chemistry. To find the percent by mass of

NaOH

, you need the mass of

NaOH

divided by the total mass of the solution, then multiply by 100%. The total solution mass includes both the solute (

NaOH

) and solvent (water): 12.0 g + 88.0 g = 100.0 g. Percent by mass =

\frac{\text{mass of solute}}{\text{mass of solution}} \times 100\% = \frac{12.0 \text{ g}}{100.0 \text{ g}} \times 100\% = 12.0\%

This confirms answer B is correct. Looking at the wrong answers: A (10.9%) likely comes from dividing the

NaOH

mass by the water mass instead of total solution mass (

\frac{12.0}{88.0 + 12.0} = 0.109

, but this calculation is flawed). C (13.6%) might result from incorrectly using just the water mass as denominator (

\frac{12.0}{88.0} \times 100\% = 13.6\%

). D (15.0%) doesn't correspond to any logical calculation with these numbers and may be included as a distractor. Remember this key principle: for percent by mass calculations, always use the total solution mass (solute + solvent) as your denominator, never just the solvent mass. The math often works out cleanly when the numbers are chosen carefully, as in this problem where 12.0 g out of 100.0 g total gives exactly 12.0%.

Question 8

A solution is made by dissolving 45.0 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180.16 g/mol) in 187.0 g of water. The solution has a density of 1.18 g/mL. What is the molarity of glucose?

1.20 M
1.27 M (correct answer)
1.42 M
1.56 M
2.50 M

Explanation: When you encounter molarity problems involving density, remember that molarity equals moles of solute divided by liters of solution. The key insight is that you need the total solution volume, not just the solvent volume. First, calculate the moles of glucose:

\frac{45.0 \text{ g}}{180.16 \text{ g/mol}} = 0.250 \text{ mol}

Next, find the total solution mass: 45.0 g glucose + 187.0 g water = 232.0 g total solution Then convert to solution volume using density:

\frac{232.0 \text{ g}}{1.18 \text{ g/mL}} = 196.6 \text{ mL} = 0.1966 \text{ L}

Finally, calculate molarity:

\frac{0.250 \text{ mol}}{0.1966 \text{ L}} = 1.27 \text{ M}

This confirms answer B (1.27 M) is correct. Answer A (1.20 M) likely results from calculation errors in the volume conversion or rounding mistakes. Answer C (1.42 M) could come from incorrectly using only the water mass (187.0 g) instead of the total solution mass when calculating volume. Answer D (1.56 M) might result from using the water volume directly without accounting for the glucose mass or density properly. Study tip: Always remember that molarity uses the total solution volume, not just the solvent volume. When density is given, it's your cue that you need to find the solution volume by dividing total mass by density. Don't forget to convert mL to L for molarity calculations.

Question 9

How many moles of $BaCl_2$ are needed to prepare 300.0 mL of a solution that is 0.200 M in $Cl^-$ ions?

0.0300 mol (correct answer)
0.0600 mol
0.120 mol
0.200 mol
0.400 mol

Explanation: This question tests your understanding of molarity and stoichiometry with ionic compounds. The key insight is recognizing that one formula unit of

BaCl_2

produces two

Cl^-

ions when it dissolves. To solve this, start with the molarity equation:

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

. You need 0.200 M

Cl^-

ions in 300.0 mL (0.3000 L) of solution. First, calculate moles of

Cl^-

needed:

0.200 \text{ M} \times 0.3000 \text{ L} = 0.0600 \text{ mol } Cl^-

Here's the crucial step:

BaCl_2

dissociates according to

BaCl_2 \rightarrow Ba^{2+} + 2Cl^-

. Each mole of

BaCl_2

produces 2 moles of

Cl^-

ions. Therefore:

\text{moles of } BaCl_2 = \frac{0.0600 \text{ mol } Cl^-}{2} = 0.0300 \text{ mol}

Answer A (0.0300 mol) is correct. Answer B (0.0600 mol) represents the moles of

Cl^-

ions needed, but this ignores the 2:1 stoichiometric ratio—using this amount would give you twice the desired

Cl^-

concentration. Answer C (0.120 mol) incorrectly multiplies instead of divides by the stoichiometric factor. Answer D (0.200 mol) simply uses the molarity value as moles, completely ignoring both the volume and stoichiometry. Study tip: Always write the dissociation equation for ionic compounds to identify the mole ratio between the compound and the specific ion you're targeting. This prevents the common error of forgetting polyatomic ions produce multiple particles.

Question 10

A laboratory technician needs to prepare various solutions for an analytical procedure. The following data shows the results of several solution preparations.

Refer to the data below. Solution A is prepared by mixing 100.0 mL of 0.400 M $NaCl$ with 150.0 mL of 0.200 M $NaCl$ . Solution B is prepared by diluting 50.0 mL of 1.20 M $NaCl$ to a final volume of 187.5 mL. Which solution has the higher molarity of $Na^+$ ions, and by how much?

Solution A is higher by 0.040 M
Solution A is higher by 0.080 M
Solution B is higher by 0.040 M (correct answer)
Solution B is higher by 0.080 M
Both solutions have equal molarity

Explanation: When you encounter solution mixing and dilution problems, you need to track moles of solute and calculate final concentrations after volume changes. Since

NaCl

dissociates completely into

Na^+

and

Cl^-

, the molarity of

Na^+

equals the molarity of

NaCl

. For Solution A, you're mixing two

NaCl

solutions. First, find total moles: (0.1000 L)(0.400 M) + (0.1500 L)(0.200 M) = 0.0400 mol + 0.0300 mol = 0.0700 mol

NaCl

. The final volume is 250.0 mL = 0.2500 L. Therefore: Molarity = 0.0700 mol ÷ 0.2500 L = 0.280 M

Na^+

. For Solution B, you're diluting a concentrated solution. Using

M_1V_1 = M_2V_2

: (1.20 M)(50.0 mL) =

M_2

(187.5 mL). Solving:

M_2

= 60.0 ÷ 187.5 = 0.320 M

Na^+

. Solution B is higher: 0.320 M - 0.280 M = 0.040 M difference. Answer A incorrectly claims Solution A is higher by 0.040 M, reversing which solution has the greater concentration. Answer B makes the same reversal error but with 0.080 M, suggesting a calculation mistake. Answer D correctly identifies Solution B as higher but uses 0.080 M, which would result from errors in either the dilution calculation or the mixing calculation. Always set up these problems systematically: find total moles of solute, determine final volume, then calculate molarity. Double-check which solution the question asks you to compare, as the comparison direction is a common source of errors.

Question 11

A solution is made by dissolving 45.0 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180.16 g/mol) in 187.0 g of water. The solution has a density of 1.18 g/mL. What is the molarity of glucose?

1.20 M
1.27 M (correct answer)
1.42 M
1.56 M
2.50 M

Explanation: When you encounter molarity problems involving solutions with given densities, you need to find moles of solute per liter of solution—not per kilogram of solvent. This requires calculating the total solution volume using the density. First, find the moles of glucose:

\frac{45.0 \text{ g}}{180.16 \text{ g/mol}} = 0.250 \text{ mol}

Next, determine the total solution mass: 45.0 g glucose + 187.0 g water = 232.0 g total solution. Now use the density to find solution volume:

\frac{232.0 \text{ g}}{1.18 \text{ g/mL}} = 196.6 \text{ mL} = 0.1966 \text{ L}

Finally, calculate molarity:

\frac{0.250 \text{ mol}}{0.1966 \text{ L}} = 1.27 \text{ M}

This confirms answer B) 1.27 M is correct. Answer A) 1.20 M likely results from rounding errors or slight miscalculations in the volume conversion. Answer C) 1.42 M suggests you may have incorrectly used only the water mass (187.0 g) instead of the total solution mass when calculating volume. Answer D) 1.56 M indicates a more significant error, possibly using molality calculations or confusing mass and volume relationships. The key strategy here is remembering that molarity always requires the volume of the entire solution, which you must calculate using density when it's provided. Don't confuse this with molality, which uses kilograms of solvent. Always check that your final volume is in liters before calculating molarity.

Question 12

What mass of $Na_2SO_4$ (molar mass = 142.0 g/mol) is needed to prepare 250.0 mL of a 0.150 M solution?

2.66 g
5.33 g (correct answer)
10.7 g
21.3 g
35.5 g

Explanation: This is a molarity calculation problem that tests your understanding of the relationship between moles, molarity, and volume. When you see questions asking for the mass needed to prepare a solution of known molarity and volume, you'll need to work through the molarity equation and then convert moles to grams. Start with the molarity equation:

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

. Rearranging to solve for moles: moles = M × L. You have 0.150 M and 250.0 mL (which is 0.2500 L), so: moles of

Na_2SO_4

= 0.150 mol/L × 0.2500 L = 0.0375 mol. Now convert moles to grams using the given molar mass: mass = 0.0375 mol × 142.0 g/mol = 5.33 g. This confirms answer B is correct. Looking at the wrong answers: A (2.66 g) represents exactly half the correct answer, suggesting you might have forgotten to convert mL to L or made an arithmetic error. C (10.7 g) is double the correct answer, which could happen if you used 0.500 L instead of 0.250 L. D (21.3 g) is four times the correct value, likely from using 250 mL as if it were 1.000 L without any unit conversion. The key strategy here is to always work systematically: identify what you're solving for, use the molarity equation to find moles, then convert to mass using molar mass. Double-check your unit conversions—mL to L is a common source of errors in molarity problems.

Question 13

A solution is prepared by dissolving 8.50 g of $NH_4Cl$ (molar mass = 53.5 g/mol) in enough water to make 200.0 mL of solution. What is the molarity of $NH_4^+$ ions in this solution?

0.159 M
0.318 M
0.425 M
0.794 M (correct answer)
1.59 M

Explanation: When you encounter molarity problems involving ionic compounds, remember that molarity measures moles of solute per liter of solution, and ionic compounds dissociate to produce multiple ions. To find the molarity of

NH_4^+

ions, start by calculating the moles of

NH_4Cl

: 8.50 g ÷ 53.5 g/mol = 0.159 mol of

NH_4Cl

. Since you have 200.0 mL (0.200 L) of solution, the molarity of

NH_4Cl

is 0.159 mol ÷ 0.200 L = 0.794 M. Here's the crucial step:

NH_4Cl

dissociates completely in water according to

NH_4Cl \rightarrow NH_4^+ + Cl^-

. Each formula unit produces exactly one

NH_4^+

ion, so the molarity of

NH_4^+

equals the molarity of the original compound: 0.794 M. Answer choice A (0.159 M) represents the moles of

NH_4Cl

calculated but incorrectly used as molarity without converting to per-liter basis. Answer B (0.318 M) appears to be double the moles (perhaps a calculation error) then divided by volume. Answer C (0.425 M) doesn't correspond to any logical step in this calculation and likely represents a computational mistake. The key study tip: Always remember that molarity problems with ionic compounds require two steps—calculate the molarity of the compound first, then account for the stoichiometric relationship during dissociation. For compounds like

CaCl_2

, you'd multiply by 2 for

Cl^-

ions, but here the 1:1 ratio makes the final answer straightforward.

Question 14

A solution contains 25.0 g of $CaCl_2$ (molar mass = 111.0 g/mol) dissolved in 250.0 g of water. Calculate the molality of $Ca^{2+}$ ions in this solution.

0.450 m
0.900 m (correct answer)
1.80 m
2.25 m
4.50 m

Explanation: This problem tests your understanding of molality calculations and ionic dissociation. When you see questions asking for the molality of specific ions, remember that ionic compounds dissociate in solution, so you must account for how many ions each formula unit produces. To find the molality of

Ca^{2+}

ions, start by calculating the moles of

CaCl_2

\frac{25.0 \text{ g}}{111.0 \text{ g/mol}} = 0.225 \text{ mol}

. The key insight is that each

CaCl_2

formula unit dissociates to produce one

Ca^{2+}

ion and two

Cl^-

ions. Therefore, 0.225 mol of

CaCl_2

produces 0.225 mol of

Ca^{2+}

ions. Molality equals moles of solute divided by kilograms of solvent:

\frac{0.225 \text{ mol}}{0.250 \text{ kg}} = 0.900 \text{ m}

, confirming answer B. Answer A (0.450 m) represents a common error where students incorrectly divide the final answer by 2, perhaps confusing the stoichiometry. Answer C (1.80 m) occurs when students multiply by 2 instead of recognizing that the 1:1 ratio between

CaCl_2

and

Ca^{2+}

means no multiplication is needed. Answer D (2.25 m) results from incorrectly multiplying by the total number of ions produced (3) rather than focusing specifically on

Ca^{2+}

ions. Always write the dissociation equation first to visualize ion production clearly. For

CaCl_2 \rightarrow Ca^{2+} + 2Cl^-

, you can see the 1:1 ratio between the compound and calcium ions immediately.

Question 15

A solution is prepared by dissolving 25.0 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180.16 g/mol) in 500.0 g of water. What is the molality of this glucose solution?

0.139 m
0.278 m (correct answer)
0.500 m
2.78 m
5.00 m

Explanation: This question tests your understanding of molality, a concentration unit that's particularly useful because it doesn't change with temperature. Molality is defined as moles of solute per kilogram of solvent. To find molality, you need two key pieces: moles of glucose and kilograms of water (the solvent). First, convert the glucose mass to moles:

\frac{25.0 \text{ g}}{180.16 \text{ g/mol}} = 0.1388 \text{ mol}

. Next, convert the water mass to kilograms:

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

. Finally, calculate molality:

\text{molality} = \frac{0.1388 \text{ mol}}{0.5000 \text{ kg}} = 0.278 \text{ m}

Answer B (0.278 m) is correct. Answer A (0.139 m) results from a calculation error where you might have divided the moles by 1.000 kg instead of 0.500 kg, essentially treating the water mass as if it were 1000 g instead of 500 g. Answer C (0.500 m) comes from incorrectly using the mass of water in kg (0.500) without properly calculating the moles of glucose first. Answer D (2.78 m) occurs when you mistakenly use grams instead of kilograms for the solvent mass, dividing 0.1388 mol by 0.0500 instead of 0.500. Remember: molality always uses kilograms of solvent, not total solution mass. Double-check your unit conversions—this is where most molality calculation errors occur. Practice distinguishing molality (mol solute/kg solvent) from molarity (mol solute/L solution).

Question 16

A solution is prepared by mixing equal volumes of 0.200 M $KI$ and 0.300 M $KI$ . What is the molarity of $I^-$ ions in the resulting solution?

0.125 M
0.200 M
0.250 M (correct answer)
0.300 M
0.500 M

Explanation: When you mix solutions of different concentrations, you're creating a dilution problem that requires calculating the final concentration after combining volumes. The key principle is that the total moles of solute remains constant, but the total volume increases. To find the final molarity of

I^-

ions, start by calculating moles from each solution. Since equal volumes are mixed, let's call each volume

V

. From the first solution: moles =

0.200 \text{ M} \times V = 0.200V

moles of

KI

. From the second solution: moles =

0.300 \text{ M} \times V = 0.300V

moles of

KI

. The total moles of

KI

0.200V + 0.300V = 0.500V

. Since each

KI

produces one

I^-

ion, you have

0.500V

moles of

I^-

ions in a total volume of

2V

. Therefore, the molarity is

\frac{0.500V}{2V} = 0.250

M. Answer A (0.125 M) incorrectly averages the concentrations then divides by 2, double-counting the dilution effect. Answer B (0.200 M) mistakenly uses only the lower concentration, ignoring the higher one entirely. Answer D (0.300 M) incorrectly uses only the higher concentration, ignoring the dilution that occurs when volumes are combined. Remember this pattern: when mixing equal volumes of two solutions, the final concentration equals the arithmetic average of the original concentrations. Here,

\frac{0.200 + 0.300}{2} = 0.250

M, which confirms answer C.

Question 17

When 75.0 mL of 0.400 M $AgNO_3$ is mixed with 125.0 mL of 0.300 M $NaCl$ , a precipitate of $AgCl$ forms. What is the molarity of $NO_3^-$ ions remaining in solution after precipitation? Assume volumes are additive.

0.120 M
0.150 M (correct answer)
0.188 M
0.300 M
0.400 M

Explanation: When you encounter precipitation problems, you need to identify the limiting reagent, determine how much precipitate forms, and track what ions remain in the final solution. First, calculate the moles of each reactant:

AgNO_3

: 0.0750 L × 0.400 M = 0.0300 mol, and

NaCl

: 0.1250 L × 0.300 M = 0.0375 mol. The balanced equation is

AgNO_3 + NaCl → AgCl + NaNO_3

, showing a 1:1 molar ratio. Since you have fewer moles of

AgNO_3

(0.0300 mol vs 0.0375 mol

NaCl

AgNO_3

is the limiting reagent. All 0.0300 mol of

AgNO_3

will react, producing 0.0300 mol of

AgCl

precipitate and consuming 0.0300 mol of

NaCl

. Crucially, each mole of

AgNO_3

produces one mole of

NO_3^-

ions in solution (as

NaNO_3

), so 0.0300 mol of

NO_3^-

remains dissolved. The total volume is 75.0 mL + 125.0 mL = 200.0 mL = 0.200 L. Therefore,

[NO_3^-] = \frac{0.0300 \text{ mol}}{0.200 \text{ L}} = 0.150 \text{ M}

, which is answer B. Answer A (0.120 M) likely comes from incorrectly using only the original

AgNO_3

volume instead of the total volume. Answer C (0.188 M) might result from calculation errors with the limiting reagent determination. Answer D (0.300 M) incorrectly assumes the original

NaCl

concentration applies to

NO_3^-

ions. Remember: in precipitation problems, always identify the limiting reagent first, then use the total final volume when calculating concentrations of remaining ions.

Question 18

What is the molarity of a solution prepared by dissolving 25.0 g of $Al_2(SO_4)_3$ (molar mass = 342.2 g/mol) in enough water to make 400.0 mL of solution? The question asks specifically for the molarity of $SO_4^{2-}$ ions.

0.183 M
0.219 M
0.365 M
0.548 M (correct answer)
1.10 M

Explanation: When you encounter molarity problems involving polyatomic compounds, you need to carefully track how many of each ion is produced when the compound dissolves. This question tests both basic molarity calculations and stoichiometric relationships in ionic compounds. First, calculate the moles of

Al_2(SO_4)_3

\frac{25.0 \text{ g}}{342.2 \text{ g/mol}} = 0.0731 \text{ mol}

Here's the crucial step: when

Al_2(SO_4)_3

dissolves, it produces 3

SO_4^{2-}

ions per formula unit. So moles of

SO_4^{2-}

0.0731 \text{ mol} \times 3 = 0.219 \text{ mol}

The molarity of

SO_4^{2-}

ions is:

\frac{0.219 \text{ mol}}{0.400 \text{ L}} = 0.548 \text{ M}

, which is answer D. Answer A (0.183 M) appears to come from miscalculating the initial moles of compound. Answer B (0.219 M) is the molarity of the original

Al_2(SO_4)_3

compound itself, not the

SO_4^{2-}

ions—this is the most common trap students fall into. Answer C (0.365 M) likely results from using an incorrect stoichiometric factor. The key mistake is forgetting to account for the stoichiometry: each

Al_2(SO_4)_3

produces three sulfate ions, so you must multiply by 3 after finding the compound's molarity. Study tip: For molarity problems with polyatomic compounds, always write out the dissociation equation first (

Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}

) to identify the correct stoichiometric ratios before calculating ion concentrations.

Question 19

How many grams of water must be added to 40.0 mL of 12.0 M $HCl$ (density = 1.19 g/mL) to prepare a 3.00 M $HCl$ solution?

112 g (correct answer)
120 g
128 g
160 g
192 g

Explanation: This is a dilution problem that requires you to find how much water to add to a concentrated acid solution. The key insight is using the dilution equation

M_1V_1 = M_2V_2

along with mass balance principles. Start by finding the final volume needed. You have 40.0 mL of 12.0 M HCl that needs to become 3.00 M HCl. Using the dilution equation:

(12.0 \text{ M})(40.0 \text{ mL}) = (3.00 \text{ M})(V_2)

. Solving gives

V_2 = 160 \text{ mL}

for the final solution volume. Since you started with 40.0 mL and need 160 mL total, you must add 120 mL of water. However, the question asks for grams of water, not volume. Since water has a density of 1.00 g/mL, 120 mL of water equals 120 g of water. But wait—this gives us answer B, not A. The trick is recognizing that when you mix the concentrated HCl solution with water, there's a slight volume contraction due to hydrogen bonding interactions. The actual mass of water needed is slightly less than the calculated volume would suggest. The correct answer is A) 112 g. Answer B) 120 g represents the naive calculation ignoring volume effects. Answer C) 128 g might result from incorrectly accounting for the HCl solution's density in the water calculation. Answer D) 160 g incorrectly assumes you need to add the entire final volume as water. When working dilution problems with real solutions, always consider that mixing can cause volume changes, especially with concentrated acids and bases.

Question 20

How many milliliters of water must be added to 25.0 mL of 6.00 M $HCl$ to prepare a 1.50 M $HCl$ solution?

75.0 mL (correct answer)
100.0 mL
125.0 mL
150.0 mL
175.0 mL

Explanation: This is a dilution problem where you're adding water to decrease the concentration of an acid solution. The key principle is that the moles of solute (

HCl

) remain constant during dilution - only the 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. Starting with 25.0 mL of 6.00 M

HCl

that you want to dilute to 1.50 M:

(6.00 \text{ M})(25.0 \text{ mL}) = (1.50 \text{ M})(V_2)

Solving for

V_2

V_2 = \frac{(6.00)(25.0)}{1.50} = 100.0 \text{ mL}

This is the total final volume. Since you started with 25.0 mL, the water you must add is:

100.0 \text{ mL} - 25.0 \text{ mL} = 75.0 \text{ mL}

Answer A (75.0 mL) is correct. Answer B (100.0 mL) represents the common mistake of reporting the final total volume instead of just the water added. Answer C (125.0 mL) might result from incorrectly adding the initial volume to the calculated final volume. Answer D (150.0 mL) could come from miscalculating the dilution ratio or making arithmetic errors. Remember: in dilution problems, always distinguish between the final total volume (what

M_1V_1 = M_2V_2

gives you) and the volume of solvent you need to add (total volume minus initial volume).