Health Education Systems Inc (HESI) A2 Exam Quiz: Solutions And Concentration

20 questions · exam conditions

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Solutions And ConcentrationQuestion 1 of 20

A chemist adds 40 g of potassium nitrate (KNO₃) to 100 g of water at 60°C. After stirring, 5 g of undissolved solid KNO₃ remains at the bottom of the beaker. Which term best describes the solution?

Unsaturated

Supersaturated

Saturated

Concentrated

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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: Solutions And Concentration

Practice Solutions And Concentration in Health Education Systems Inc (HESI) A2 Exam 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 Concentration, giving you a quick way to practice the rules, question types, and explanations that matter most for Health Education Systems Inc (HESI) A2 Exam.

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

Unsaturated
Supersaturated
Saturated (correct answer)
Concentrated

Explanation: When you encounter solubility questions, focus on the relationship between what dissolves and what remains undissolved to determine the solution's saturation state. In this scenario, you added 40 g of KNO₃ to water, but only 35 g dissolved (since 5 g remained undissolved). This tells you that at 60°C, the water can hold exactly 35 g of KNO₃ before reaching its maximum capacity. Since undissolved solid remains in equilibrium with the dissolved solute, the solution has reached its saturation point—it contains the maximum amount of solute that can dissolve at that temperature and pressure. Answer C (Saturated) correctly describes this state. A saturated solution exists when no more solute can dissolve, and any additional solute remains as undissolved solid at the bottom. Answer A (Unsaturated) is wrong because an unsaturated solution could dissolve more solute—there wouldn't be any undissolved solid remaining. Answer B (Supersaturated) is incorrect because supersaturated solutions contain more dissolved solute than normally possible at that temperature, typically created by cooling a hot saturated solution. These solutions are unstable and usually don't have undissolved solid present initially. Answer D (Concentrated) describes the relative amount of solute but doesn't indicate saturation status—concentrated solutions can be saturated or unsaturated. Remember this key indicator: if undissolved solid remains after stirring and the solution is at equilibrium, you're looking at a saturated solution. The presence of undissolved solid is your clearest clue that the solution has reached its solubility limit.

Question 2

A laboratory stock solution of hydrochloric acid (HCl) is 12.0 M. How much of this stock solution is required to prepare 400.0 mL of a 3.0 M HCl solution?

100.0 mL (correct answer)
133.3 mL
300.0 mL
1600.0 mL

Explanation: When you encounter dilution problems, you're working with the principle that the amount of solute (moles) remains constant while the solvent changes. This is a classic application of the dilution formula:

M_1V_1 = M_2V_2

, where the subscripts represent initial and final conditions. Here, you need to find how much 12.0 M stock solution is required to make 400.0 mL of 3.0 M solution. Setting up the equation:

(12.0 \text{ M})(V_1) = (3.0 \text{ M})(400.0 \text{ mL})

Solving for

V_1

V_1 = \frac{(3.0)(400.0)}{12.0} = \frac{1200}{12.0} = 100.0 \text{ mL}

Choice A (100.0 mL) is correct because this amount of concentrated stock, when diluted to 400.0 mL total volume, gives the desired 3.0 M concentration. Choice B (133.3 mL) results from incorrectly dividing 400.0 by 3.0, ignoring the relationship between initial and final concentrations. Choice C (300.0 mL) comes from subtracting the final concentration from the final volume (400.0 - 100.0), which has no chemical basis. Choice D (1600.0 mL) appears to multiply concentrations and volume incorrectly, perhaps calculating (12.0 × 3.0 × 400.0)/900. Remember the key principle: you're always diluting from higher to lower concentration, so you'll need less volume of the concentrated solution than your final volume. If your calculated stock volume exceeds the final solution volume, double-check your setup—this is physically impossible in dilution problems.

Question 3

How many grams of glucose (C₆H₁₂O₆, molar mass ≈ 180.16 g/mol) are required to prepare 200.0 mL of a 0.25 M solution?

4.5 g
9.0 g (correct answer)
45.0 g
90.1 g

Explanation: When you encounter molarity problems, you're working with the relationship between moles of solute, volume of solution, and concentration. The key formula is:

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

To find the grams of glucose needed, first calculate the moles required. Rearranging the molarity equation: moles = Molarity × Volume (in L). Convert 200.0 mL to liters: 200.0 mL = 0.2000 L. Then: moles of glucose = 0.25 M × 0.2000 L = 0.050 mol. Next, convert moles to grams using the molar mass: grams = moles × molar mass = 0.050 mol × 180.16 g/mol = 9.0 g. Looking at the wrong answers: Choice A (4.5 g) results from forgetting to convert mL to L, using 200 instead of 0.2000 in the calculation. Choice C (45.0 g) comes from incorrectly using 200 mL as if it were 2.0 L instead of 0.2 L. Choice D (90.1 g) represents using the full 200 mL as liters without any conversion. The correct answer is B (9.0 g). For HESI success, always write out your unit conversions explicitly in molarity problems. The most common mistakes involve volume unit errors—mL versus L confusion accounts for most wrong answers. Create a systematic approach: identify what you're solving for, convert units first, then plug into formulas. This prevents the calculation errors that lead to attractive but incorrect answer choices.

Question 4

A student prepares a solution by dissolving 29.22 grams of sodium chloride (NaCl, molar mass ≈ 58.44 g/mol) in enough water to make a final volume of 500.0 mL. What is the molarity (M) of the resulting solution?

0.25 M
0.50 M
1.00 M (correct answer)
2.00 M

Explanation: When you encounter molarity calculations, you're working with one of chemistry's most fundamental concentration units. Molarity (M) equals moles of solute divided by liters of solution, so you need to convert grams to moles and milliliters to liters. Start by finding moles of NaCl:

\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{29.22 \text{ g}}{58.44 \text{ g/mol}} = 0.500 \text{ mol}

Next, convert volume to liters:

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

Now calculate molarity:

M = \frac{0.500 \text{ mol}}{0.5000 \text{ L}} = 1.00 \text{ M}

This confirms answer C is correct. Looking at the wrong answers: A (0.25 M) results from incorrectly using 2.0 L as the volume instead of 0.5 L—a common error when students confuse the conversion factor. B (0.50 M) happens if you forget to convert mL to L entirely, dividing 0.500 mol by 500 instead of 0.5. D (2.00 M) occurs when students flip the calculation, dividing liters by moles instead of moles by liters. For HESI success, always follow the same systematic approach: convert grams to moles using molar mass, convert mL to L by dividing by 1000, then apply the molarity formula. Double-check your unit conversions—volume mistakes are the most common trap in concentration problems.

Question 5

The addition of a non-volatile solute, such as salt, to a pure solvent, such as water, will cause which of the following changes to the solvent's properties?

The boiling point will decrease and the freezing point will increase.
The boiling point will increase and the freezing point will decrease. (correct answer)
Both the boiling point and the freezing point will increase.
Both the boiling point and the freezing point will decrease.

Explanation: When you encounter questions about adding solutes to solvents, you're dealing with colligative properties - properties that depend on the number of particles in solution, not their identity. These are fundamental concepts in chemistry that appear regularly on the HESI. Adding a non-volatile solute like salt to water disrupts the solvent's normal behavior in two key ways. First, it elevates the boiling point because the solute particles interfere with water molecules' ability to escape into the gas phase - more energy (higher temperature) is needed to overcome these interactions. Second, it depresses the freezing point because the solute particles disrupt the orderly crystal formation required for freezing, requiring a lower temperature to achieve the solid state. Looking at the answer choices: Choice A incorrectly suggests the boiling point decreases and freezing point increases - this reverses both effects. Choice C claims both boiling and freezing points increase, which misses that freezing point actually decreases. Choice D states both decrease, which gets the freezing point direction right but wrong for boiling point. Choice B correctly identifies that the boiling point increases while the freezing point decreases. This is why we add salt to pasta water (raises boiling point for faster cooking) and to icy roads in winter (lowers freezing point to melt ice). For HESI success, remember the phrase "boiling up, freezing down" when solutes are added. Questions about colligative properties often test whether you can distinguish between the opposite effects on these two phase transitions.

Question 6

Consider two beakers of aqueous NaCl solution. Beaker A contains 100 mL of a 2.0 M solution. Beaker B contains 200 mL of a 1.5 M solution. Which beaker contains a greater number of moles of NaCl?

Beaker A contains more moles of NaCl.
Beaker B contains more moles of NaCl. (correct answer)
Both beakers contain the same number of moles of NaCl.
It is impossible to determine without knowing the mass of NaCl used.

Explanation: When you encounter molarity problems, remember that molarity (M) equals moles of solute divided by liters of solution. To find the actual number of moles, you rearrange this relationship: moles = molarity × volume in liters. Let's calculate the moles in each beaker. For Beaker A:

\text{moles} = 2.0 \text{ M} \times 0.100 \text{ L} = 0.20 \text{ mol NaCl}

For Beaker B:

\text{moles} = 1.5 \text{ M} \times 0.200 \text{ L} = 0.30 \text{ mol NaCl}

Beaker B contains more moles of NaCl (0.30 mol vs. 0.20 mol), making answer B correct. Answer A is incorrect because even though Beaker A has a higher molarity (2.0 M vs. 1.5 M), its smaller volume means fewer total moles. This is a common trap—students sometimes assume higher concentration always means more substance. Answer C is wrong because the calculations clearly show different amounts: 0.20 mol versus 0.30 mol. Answer D represents a fundamental misunderstanding. You don't need the mass of NaCl because molarity already tells you the moles per liter. The molarity and volume provide all the information needed to calculate moles directly. Remember this key insight for the HESI: molarity problems often test whether you understand that concentration and total amount are different concepts. A less concentrated solution can still contain more total solute if the volume is sufficiently larger. Always calculate the actual moles rather than just comparing concentrations.

Question 7

A student carefully prepares a saturated solution of sodium acetate at a high temperature and then allows it to cool slowly without disturbance. The resulting solution contains more dissolved solute than a saturated solution at the cooler temperature. This type of solution is best described as:

an ideal solution
an unsaturated solution
a heterogeneous mixture
a supersaturated solution (correct answer)

Explanation: When you encounter questions about solubility and solution types, focus on the relationship between temperature, dissolved solute amount, and equilibrium conditions. This scenario describes a classic supersaturated solution formation. The student created a saturated solution at high temperature (where sodium acetate has high solubility), then cooled it slowly without disturbance. As temperature decreased, the solubility of sodium acetate also decreased, but the excess solute remained dissolved instead of precipitating out. This creates an unstable solution containing more dissolved solute than would normally be possible at the lower temperature. Answer D is correct because a supersaturated solution is defined as one containing more dissolved solute than a saturated solution can hold at that specific temperature. These solutions are metastable and will precipitate excess solute if disturbed. Answer A is wrong because an ideal solution refers to one that follows Raoult's law perfectly, which relates to vapor pressure behavior, not solubility limits. Answer B is incorrect because an unsaturated solution contains less solute than the maximum amount that can dissolve at that temperature—the opposite of what's described here. Answer C is wrong because the solution remains homogeneous (single phase) even though it's supersaturated; you can't see the excess solute since it's still dissolved. Remember this pattern: when you see slow cooling of a hot saturated solution without disturbance, think supersaturation. The key clue is "more dissolved solute than a saturated solution at the cooler temperature"—this phrase directly defines supersaturation.

Question 8

What is the defining characteristic of an aqueous solution?

The solute must be a solid ionic compound.
The solution must be able to conduct electricity.
The solvent is always the substance water (H₂O). (correct answer)
The solution must have a neutral pH of 7.0.

Explanation: When you encounter questions about solution types, focus on the fundamental definition that distinguishes each category. An aqueous solution is defined by its solvent, not by the properties of what's dissolved in it. The correct answer is C because "aqueous" literally means "containing water" or "dissolved in water." By definition, any solution where water (H₂O) serves as the solvent is an aqueous solution. This is a terminology-based question where the name itself provides the key—think of "aqua" meaning water. Let's examine why the other options are incorrect. Option A is wrong because aqueous solutions can contain any type of solute—gases like CO₂ in soda, liquids like ethanol in alcoholic beverages, or molecular compounds like sugar, not just solid ionic compounds. Option B incorrectly assumes all aqueous solutions conduct electricity, but pure water with non-ionic solutes (like sugar water) won't conduct electricity since no ions are present. Option D is false because aqueous solutions can have any pH value—lemon juice (acidic, pH ~2) and ammonia solutions (basic, pH ~11) are both aqueous solutions despite being far from neutral. For HESI chemistry questions, remember that solution classifications are based on the solvent used. Aqueous means water is the solvent, regardless of what's dissolved or the solution's resulting properties. Don't get distracted by the solution's behavior—focus on its composition.

Question 9

A solution is prepared by mixing 70 mL of pure isopropyl alcohol with enough water to make a final volume of 100 mL. What is the concentration of this solution, expressed as a volume/volume percent (% v/v)?

30% v/v
70% v/v (correct answer)
100% v/v
170% v/v

Explanation: When you encounter concentration problems on the HESI, you're working with the fundamental relationship between solute, solvent, and total solution volume. Volume/volume percent (% v/v) specifically measures how much solute volume exists in the total solution volume. The formula for % v/v is:

\frac{\text{volume of solute}}{\text{total volume of solution}} \times 100\%

In this problem, you have 70 mL of pure isopropyl alcohol (the solute) mixed with enough water to create a final solution volume of 100 mL. Plugging into the formula:

\frac{70 \text{ mL}}{100 \text{ mL}} \times 100\% = 70\% \text{ v/v}

This confirms answer choice B is correct. Looking at the wrong answers: A (30% v/v) represents a common error where students might subtract the alcohol volume from the total volume and use that as the numerator—this would give you the water percentage, not the alcohol percentage. C (100% v/v) would only be correct if you had pure alcohol with no dilution. D (170% v/v) suggests adding the alcohol volume to the total volume, which misunderstands that the 100 mL already includes the alcohol. Remember this key distinction: the denominator in % v/v is always the final total volume of the mixed solution, not the volume of solvent added. Many HESI concentration problems test whether you can identify what constitutes the "total volume" versus individual component volumes.

Question 10

A nurse needs to prepare 250 mL of a 0.15 M glucose solution from a stock solution of 1.8 M glucose. After calculating the required volume of stock solution, the nurse accidentally uses twice that amount. What is the molarity of the resulting solution?

0.30 M (correct answer)
0.27 M
0.54 M
0.75 M

Explanation: First, calculate the required stock volume: M₁V₁ = M₂V₂, so (1.8 M)(V₁) = (0.15 M)(250 mL), giving V₁ = 20.8 mL. The nurse uses twice this amount (41.6 mL) in 250 mL total volume. Using M₁V₁ = M₂V₂: (1.8 M)(41.6 mL) = M₂(250 mL), so M₂ = 0.30 M. Choice B results from calculation errors in the dilution formula. Choice C incorrectly doubles the intended concentration. Choice D assumes incorrect stock solution usage.

Question 11

A laboratory receives a solution labeled as 500 ppm sodium chloride. If this solution is diluted 1:4 with distilled water, and then 200 mL of this diluted solution is further diluted to a final volume of 1.0 L, what is the final concentration in ppm?

12.5 ppm
100 ppm
50 ppm
25 ppm (correct answer)

Explanation: When you encounter dilution problems on the HESI, you're working with the principle that the amount of solute remains constant while the volume changes. This requires tracking concentration changes through multiple dilution steps. Start with the initial solution: 500 ppm sodium chloride. The first dilution is 1:4, meaning 1 part original solution mixed with 4 parts total volume (1 part solution + 3 parts water). This creates a 5-fold dilution, so the new concentration is

500 \text{ ppm} \div 5 = 100 \text{ ppm}

. Next, 200 mL of this 100 ppm solution is diluted to 1.0 L (1000 mL). The dilution factor is

\frac{1000 \text{ mL}}{200 \text{ mL}} = 5

. Therefore, the final concentration is

100 \text{ ppm} \div 5 = 25 \text{ ppm}

. Looking at the distractors: Choice A (12.5 ppm) results from incorrectly calculating the first dilution as 1:4 meaning 4-fold instead of 5-fold dilution. Choice B (100 ppm) stops after the first dilution step, ignoring the second dilution entirely. Choice C (50 ppm) likely comes from miscalculating one of the dilution factors, perhaps confusing the 1:4 ratio or the volume ratios. For HESI dilution problems, always work step-by-step through each dilution, clearly identifying whether ratios represent parts of solution to total volume or solution to diluent. Write out each calculation to avoid compounding errors across multiple steps.

Question 12

A quality control analyst tests a cleaning solution and finds it contains 3.5% (w/w) active ingredient. If the solution has a density of 1.12 g/mL, what is the molarity of the active ingredient, assuming its molecular weight is 74.5 g/mol?

0.470 M
0.526 M (correct answer)
0.392 M
0.615 M

Explanation: When you encounter concentration problems involving weight percent and molarity, you need to convert between different ways of expressing concentration using density as the bridge. To find molarity, you need moles of solute per liter of solution. Start with the given information: 3.5% (w/w) means 3.5 g of active ingredient per 100 g of solution, density is 1.12 g/mL, and molecular weight is 74.5 g/mol. First, convert the solution mass to volume: 100 g of solution ÷ 1.12 g/mL = 89.29 mL = 0.08929 L. Next, convert grams of active ingredient to moles: 3.5 g ÷ 74.5 g/mol = 0.04698 mol. Finally, calculate molarity: 0.04698 mol ÷ 0.08929 L = 0.526 M. Looking at the wrong answers: Choice A (0.470 M) likely results from calculation errors in the unit conversions or rounding too early. Choice C (0.392 M) might come from incorrectly using the density in the calculation, perhaps dividing when you should multiply. Choice D (0.615 M) could result from forgetting to convert mL to L, leading to an inflated molarity value. The correct answer is B (0.526 M). Remember this systematic approach for concentration conversions: use the given percentages to establish mass relationships, apply density to convert between mass and volume, then use molecular weight to convert mass to moles. Always track your units carefully—many errors in these problems stem from unit conversion mistakes rather than conceptual misunderstandings.

Question 13

A nurse prepares an oral medication by dissolving 2.4 g of drug powder in enough water to make exactly 80 mL of solution. If a patient needs to receive 15 mg of the drug per kilogram of body weight, and the patient weighs 65 kg, what volume of the prepared solution should be administered?

40.6 mL
32.5 mL (correct answer)
48.8 mL
28.1 mL

Explanation: Dosage calculation problems like this require you to work systematically through three steps: determine the total dose needed, find the concentration of your solution, and calculate the required volume. First, calculate the total dose required. The patient weighs 65 kg and needs 15 mg per kg:

65 \text{ kg} \times 15 \text{ mg/kg} = 975 \text{ mg}

Next, determine the concentration of your prepared solution. You dissolved 2.4 g of drug powder in 80 mL total volume. Convert grams to milligrams:

2.4 \text{ g} = 2,400 \text{ mg}

. The concentration is:

\frac{2,400 \text{ mg}}{80 \text{ mL}} = 30 \text{ mg/mL}

Finally, calculate the volume needed:

\frac{975 \text{ mg}}{30 \text{ mg/mL}} = 32.5 \text{ mL}

Choice A (40.6 mL) represents a calculation error, likely from incorrectly using the original 2.4 g without converting to mg or making an arithmetic mistake in the final division. Choice C (48.8 mL) suggests confusion in the dose calculation or concentration determination. Choice D (28.1 mL) appears to result from calculation errors in either the patient's total dose requirement or the solution concentration. For HESI dosage calculations, always double-check your unit conversions (grams to milligrams is crucial here) and work methodically through each step. Write down your intermediate calculations to avoid arithmetic errors, and always verify that your final answer makes logical sense given the numbers in the problem.

Question 14

An IV solution contains 0.85 g of sodium chloride in 100 mL of solution. If a patient receives 1.5 L of this solution over 8 hours, what is the total mass of sodium chloride administered, and what was the average rate of sodium chloride delivery in mg/min?

8.50 g total; 17.7 mg/min average rate
12.75 g total; 31.2 mg/min average rate
15.30 g total; 26.6 mg/min average rate
12.75 g total; 26.6 mg/min average rate (correct answer)

Explanation: When you encounter IV dosage calculations involving concentration and flow rates, break the problem into clear steps: find the total drug amount delivered, then calculate the delivery rate over time. First, determine the concentration: 0.85 g NaCl per 100 mL gives you

\frac{0.85 \text{ g}}{100 \text{ mL}} = 0.0085 \text{ g/mL}

. Next, calculate total sodium chloride delivered: 1.5 L = 1,500 mL, so

1,500 \text{ mL} \times 0.0085 \text{ g/mL} = 12.75 \text{ g}

total. For the delivery rate, convert the total mass to mg (12.75 g = 12,750 mg) and the time to minutes (8 hours = 480 minutes). The average rate is

\frac{12,750 \text{ mg}}{480 \text{ min}} = 26.6 \text{ mg/min}

. Choice A incorrectly calculates the total mass as if only 1 L was administered instead of 1.5 L, giving 8.50 g, then compounds this error in the rate calculation. Choice B gets the correct total mass of 12.75 g but makes an error in time conversion, likely using 6.8 hours instead of 8 hours, resulting in 31.2 mg/min. Choice C uses an incorrect concentration calculation, possibly confusing the setup, leading to 15.30 g total mass. Choice D provides both correct values: 12.75 g total and 26.6 mg/min average rate. Remember to always convert units systematically and double-check your arithmetic. HESI dosage questions frequently test unit conversion skills alongside basic proportional reasoning, so practice converting between g/mg and hours/minutes until it becomes automatic.

Question 15

A patient's blood glucose level is reported as 180 mg/dL. If the molecular weight of glucose is 180 g/mol, and assuming blood density is 1.0 g/mL, what is the approximate molarity of glucose in the patient's blood?

0.010 M (correct answer)
0.001 M
0.100 M
1.000 M

Explanation: Convert mg/dL to molarity: 180 mg/dL = 0.18 g/L. Molarity = (0.18 g/L) ÷ (180 g/mol) = 0.001 mol/L = 0.001 M. Wait - this seems too low. Let me recalculate: 180 mg/dL = 1800 mg/L = 1.8 g/L. Molarity = (1.8 g/L) ÷ (180 g/mol) = 0.010 M. Choice B results from forgetting the dL to L conversion factor of 10. Choice C multiplies instead of divides by molecular weight. Choice D uses mg as grams directly.

Question 16

When a concentrated aqueous solution is diluted by adding more water, which of the following statements is correct?

The molarity of the solution increases, and the total moles of solute increase.
The molarity of the solution decreases, and the total moles of solute remain the same. (correct answer)
The molarity of the solution remains the same, but the total moles of solute decrease.
The molarity of the solution decreases, and the total moles of solute decrease.

Explanation: When you encounter dilution problems, focus on what happens to the amount of solute versus the concentration. Dilution adds solvent (water) but never adds or removes the actual dissolved substance. The correct answer is B because dilution affects molarity and moles of solute differently. Molarity is defined as

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

. When you add water to a concentrated solution, you're increasing the denominator (total volume) while keeping the numerator (moles of solute) constant. This mathematical relationship means molarity must decrease. However, the total moles of solute remain unchanged because you're only adding water – no solute particles are created or destroyed. Choice A is incorrect because while it correctly states that moles of solute remain constant, it wrongly claims molarity increases. Adding water always decreases concentration. Choice C is wrong on both counts – molarity definitely changes during dilution (it decreases), and moles of solute stay the same, not decrease. Choice D incorrectly suggests that moles of solute decrease. This would only happen if you physically removed some of the dissolved substance, which doesn't occur in simple dilution. Study tip: Remember the dilution mantra: "Same solute, more solution, lower concentration." The actual amount of dissolved substance never changes in dilution – only its concentration does. This principle applies whether you're calculating drug concentrations in healthcare or preparing laboratory solutions.

Question 17

A student observes that a carbonated beverage fizzes more rapidly when it is warmed up. Which statement best explains this phenomenon based on the principles of solubility?

The solubility of solid sugars in the beverage increases with temperature, causing fizzing.
Increasing the temperature increases the kinetic energy and solubility of dissolved gases.
The solubility of the gaseous solute (CO₂) decreases as the temperature of the solvent increases. (correct answer)
The pressure inside the beverage container increases with temperature, forcing the gas out.

Explanation: When you encounter questions about gas behavior in liquids, focus on how temperature affects gas solubility—this follows different rules than solid solubility. The fizzing occurs because carbon dioxide (CO₂) becomes less soluble in water as temperature increases. Unlike most solids, gases have an inverse relationship with temperature: higher temperatures provide more kinetic energy to gas molecules, making them more likely to escape from the liquid phase into the gas phase. When you warm a carbonated beverage, the dissolved CO₂ can no longer stay in solution as effectively, so it forms bubbles and escapes as fizz. Answer A incorrectly focuses on sugar solubility, but sugar dissolving doesn't create gas bubbles—only the CO₂ gas can cause fizzing. Answer B contains a critical error: while temperature does increase kinetic energy, it actually decreases gas solubility, not increases it. This is a common misconception since solid solubility typically increases with temperature. Answer D mentions pressure effects, but the primary driver here is temperature's direct effect on gas solubility, not pressure changes from thermal expansion. The correct answer is C because it accurately describes the fundamental principle: gas solubility decreases as solvent temperature increases. Study tip: Remember that gas solubility behaves opposite to most solid solubility—think of hot soda going flat quickly versus cold soda staying fizzy longer. On the HESI, gas solubility questions often test whether you know this inverse temperature relationship.

Question 18

Which of the following substances would be most likely to dissolve in carbon tetrachloride (CCl₄), a nonpolar solvent?

Sodium chloride (NaCl), an ionic compound
Ammonia (NH₃), a polar molecule
Methane (CH₄), a nonpolar molecule (correct answer)
Sucrose (C₁₂H₂₂O₁₁), a large polar molecule

Explanation: When you encounter solubility questions, remember the fundamental principle: "like dissolves like." Polar substances dissolve in polar solvents, while nonpolar substances dissolve in nonpolar solvents. Since carbon tetrachloride (CCl₄) is a nonpolar solvent, you need to identify which substance is also nonpolar. Methane (CH₄) is the correct answer because it's a nonpolar molecule. The carbon-hydrogen bonds have very small electronegativity differences, and the tetrahedral molecular geometry creates a symmetrical distribution of electron density. This nonpolar nature makes methane highly compatible with the nonpolar CCl₄. Option A is incorrect because sodium chloride (NaCl) is an ionic compound that dissociates into charged ions (Na⁺ and Cl⁻) in solution. These charged particles are strongly attracted to polar solvents like water, not nonpolar solvents like CCl₄. Option B is wrong because ammonia (NH₃) is polar due to nitrogen's high electronegativity and the molecule's trigonal pyramidal shape, which creates an uneven electron distribution. Polar molecules like NH₃ prefer polar solvents. Option D is incorrect because sucrose is a large polar molecule with multiple hydroxyl (-OH) groups that can form hydrogen bonds. These polar characteristics make sucrose highly soluble in water but essentially insoluble in nonpolar solvents. Study tip: For HESI chemistry questions about solubility, quickly categorize each substance as ionic, polar, or nonpolar, then match it to the solvent's polarity. This pattern recognition will save you time and improve accuracy on similar questions.

Question 19

In a solution of tincture of iodine, solid iodine (I₂) is dissolved in ethanol (C₂H₅OH), a liquid. In this solution, which substance is the solute and which is the solvent?

Iodine is the solute and ethanol is the solvent. (correct answer)
Ethanol is the solute and iodine is the solvent.
Both iodine and ethanol are solutes.
Both iodine and ethanol are solvents.

Explanation: When you encounter solution chemistry questions, focus on identifying which substance dissolves (solute) and which does the dissolving (solvent). The key principle is that the solute is typically the substance present in smaller amounts that gets dissolved, while the solvent is present in larger amounts and does the dissolving. In tincture of iodine, solid iodine crystals are dissolved into liquid ethanol. The iodine transforms from its solid state into individual molecules dispersed throughout the ethanol. Since iodine is being dissolved and is present in much smaller quantities, it's the solute. Ethanol, being the liquid that dissolves the iodine and present in much larger amounts, serves as the solvent. Looking at the wrong answers: Option B reverses the roles incorrectly—ethanol cannot be the solute when it's the liquid medium doing the dissolving. Option C suggests both substances are solutes, but this is impossible since every solution must have a solvent to dissolve the solutes into. Option D claims both are solvents, which violates the basic definition of a solution—you need something to be dissolved (solute) for a solution to exist. Remember this pattern: in most solutions you'll encounter on the HESI, liquids are typically solvents (especially water, alcohols, and other common liquid mediums) while solids, gases, or smaller amounts of other liquids are the solutes. When in doubt, ask yourself "what's doing the dissolving?" and "what's being dissolved?"

Question 20

A student adds 20.0 mL of a 2.5 M NaOH solution to a volumetric flask and adds water to reach a final volume of 100.0 mL. What is the final concentration of the NaOH solution?

0.25 M
0.50 M (correct answer)
1.0 M
12.5 M

Explanation: This question tests dilution calculations, a fundamental concept in chemistry where you're adding solvent to decrease the concentration of a solution. When you see a dilution problem, remember that the amount of solute (the substance being dissolved) stays constant—only the volume changes. Use the dilution formula:

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. Here, you have 20.0 mL of 2.5 M NaOH diluted to 100.0 mL total volume. Solving for the final concentration:

(2.5 \text{ M})(20.0 \text{ mL}) = M_2(100.0 \text{ mL})

. This gives you

M_2 = \frac{50.0}{100.0} = 0.50 \text{ M}

, which is answer B. Answer A (0.25 M) represents a common error where students might divide by 4 instead of 5, confusing the dilution factor. Answer C (1.0 M) suggests incorrectly thinking the concentration doubles rather than decreases during dilution. Answer D (12.5 M) comes from multiplying instead of using the dilution relationship, which defies the basic principle that dilution always decreases concentration. For HESI dilution problems, always identify what stays constant (moles of solute) versus what changes (volume and concentration). Set up

M_1V_1 = M_2V_2

and solve systematically. Remember: diluting always makes solutions less concentrated, so your final answer should be smaller than the original molarity.