- 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?
Answer: 32.5 mL
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 kg×15 mg/kg=975 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 g=2,400 mg. The concentration is: 80 mL2,400 mg=30 mg/mL
Finally, calculate the volume needed: 30 mg/mL975 mg=32.5 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. - 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?
Answer: 0.30 M
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.
- Which of the following statements correctly distinguishes between a concentrated solution and a saturated solution?
Answer: A solution can be concentrated but still unsaturated if the solute has very high solubility.
When you encounter questions about solution properties, you need to distinguish between concentration (how much solute is present) and saturation (whether more solute can dissolve). These are independent concepts that often confuse students.
A solution can indeed be concentrated but still unsaturated if the solute has very high solubility, making answer C correct. Consider sugar water: you can dissolve a large amount of sugar (making it concentrated) while still being able to dissolve more (keeping it unsaturated) because sugar is extremely soluble in water. The concentration depends on how much solute you've added, while saturation depends on the maximum amount that can dissolve under those conditions.
Answer A incorrectly assumes concentration and saturation are linked. A concentrated solution only means you have a lot of solute present—it says nothing about whether more could dissolve. Answer B makes the opposite error, suggesting saturated solutions must be concentrated. However, a solution can be saturated with very little solute if that solute has low solubility—like trying to dissolve salt in oil. Answer D creates a false distinction based on polarity. Both polar and nonpolar solutes can create concentrated or saturated solutions depending on the solvent and conditions.
For HESI success, remember that concentration is quantitative (how much is there) while saturation is qualitative (can more dissolve). These properties are independent—always consider the solute's inherent solubility in that specific solvent when evaluating whether a solution can be both concentrated and unsaturated.
- 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?
Answer: 1.00 M
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: moles=molar massmass=58.44 g/mol29.22 g=0.500 mol
Next, convert volume to liters: 500.0 mL=0.5000 L
Now calculate molarity: M=0.5000 L0.500 mol=1.00 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. - When a concentrated aqueous solution is diluted by adding more water, which of the following statements is correct?
Answer: The molarity of the solution decreases, and the total moles of solute remain the same.
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 Molarity=liters of solutionmoles of solute. 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. - 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?
Answer: The boiling point will increase and the freezing point will decrease.
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.
- Which of the following substances would be most likely to dissolve in carbon tetrachloride (CCl₄), a nonpolar solvent?
Answer: Methane (CH₄), a nonpolar molecule
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.
- 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?
Answer: 12.75 g total; 26.6 mg/min average rate
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 100 mL0.85 g=0.0085 g/mL. Next, calculate total sodium chloride delivered: 1.5 L = 1,500 mL, so 1,500 mL×0.0085 g/mL=12.75 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 480 min12,750 mg=26.6 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. - A nurse is preparing a 0.9% (m/v) normal saline solution. This concentration means 0.9 grams of NaCl per 100 mL of solution. How many grams of NaCl are needed to prepare 2.5 liters of this solution?
Answer: 22.5 g
When you encounter concentration problems on the HESI, you're working with ratio and proportion calculations that are fundamental to safe medication preparation. The key is understanding what the percentage concentration tells you and then scaling it up or down proportionally.
A 0.9% (m/v) solution means 0.9 grams of solute per 100 mL of solution. To find how much NaCl you need for 2.5 liters, first convert to the same units: 2.5 L = 2,500 mL. Now set up a proportion: 100 mL0.9 g=2,500 mLx g
Cross multiply: x=1000.9×2,500=1002,250=22.5 g
This confirms answer C is correct.
Looking at the wrong answers: A) 2.25 g results from incorrectly using 250 mL instead of 2,500 mL in your calculation - a decimal place error when converting liters to milliliters. B) 9.0 g comes from multiplying 0.9 g by 10, suggesting someone thought 2.5 L equals 1,000 mL instead of 2,500 mL. D) 225 g occurs when you multiply by 250 instead of 25 (the scaling factor from 100 mL to 2,500 mL).
For HESI concentration problems, always convert to matching units first, then set up your proportion carefully. Double-check your unit conversions - especially L to mL - as these are common sources of error that the test writers exploit in the distractors. - A pharmacy technician mixes 150 mL of 20% (w/v) saline solution with 350 mL of 8% (w/v) saline solution. What is the final concentration of the mixed solution?
Answer: 11.6% (w/v)
Use the dilution principle: C₁V₁ + C₂V₂ = C₃V₃. Mass from solution 1: (20%)(150 mL) = 30 g. Mass from solution 2: (8%)(350 mL) = 28 g. Total mass = 58 g, total volume = 500 mL. Final concentration = (58 g)/(500 mL) × 100% = 11.6% (w/v). Choice A incorrectly averages the concentrations without considering volumes. Choice B uses simple arithmetic mean of percentages. Choice D results from calculation errors in the mass balance.
- 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?
Answer: 0.010 M
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.
- 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?
Answer: 0.526 M
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.
- Which of the following aqueous solutions would be the best conductor of electricity?
Answer: A 0.5 M solution of potassium chloride (KCl)
When you encounter questions about electrical conductivity in aqueous solutions, think about which substances produce ions when dissolved in water. Electrical conductivity depends on the presence of mobile charged particles (ions) that can carry electric current.
Potassium chloride (KCl) is an ionic compound that completely dissociates in water: KCl→K++Cl−. This creates two ions per formula unit, providing abundant charged particles to conduct electricity. The 0.5 M KCl solution will have a high concentration of these mobile ions, making it an excellent electrical conductor.
Looking at why the other options are poor conductors: Choice A (sugar) is a covalent compound that dissolves as whole molecules without forming ions - C6H12O6 remains intact in solution. Choice C (ethanol) is also covalent and stays as complete C2H5OH molecules when dissolved, producing virtually no ions. Choice D (pure deionized water) has extremely few ions since water only very weakly ionizes (H2O⇌H++OH−), making it a poor conductor.
The answer is B because only KCl produces a significant concentration of ions in solution.
Study tip for the HESI: When evaluating electrical conductivity, immediately identify whether each substance is ionic (will dissociate into ions) or covalent (stays as molecules). Ionic compounds in solution almost always conduct electricity better than molecular compounds. Focus on recognizing common ionic compounds like salts (NaCl, KCl, CaCl₂) versus molecular substances like sugars and alcohols. - 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?
Answer: Beaker B contains more moles of NaCl.
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: moles=2.0 M×0.100 L=0.20 mol NaCl
For Beaker B: moles=1.5 M×0.200 L=0.30 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. - 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?
Answer: Saturated
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.
- 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?
Answer: 9.0 g
When you encounter molarity problems, you're working with the relationship between moles of solute, volume of solution, and concentration. The key formula is: Molarity=liters of solutionmoles of solute
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. - 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?
Answer: 0.50 M
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: M1V1=M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 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 M)(20.0 mL)=M2(100.0 mL). This gives you M2=100.050.0=0.50 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 M1V1=M2V2 and solve systematically. Remember: diluting always makes solutions less concentrated, so your final answer should be smaller than the original molarity. - What is the defining characteristic of an aqueous solution?
Answer: The solvent is always the substance water (H₂O).
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.
- 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?
Answer: The solubility of the gaseous solute (CO₂) decreases as the temperature of the solvent increases.
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.
- 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)?
Answer: 70% v/v
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: total volume of solutionvolume of solute×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: 100 mL70 mL×100%=70% 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.
