2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
Each numbered box in the diagram represents a liquid sample at the particulate level. A line joining two particles represents a chemical bond. Which box represents a solution?
Answer and reasoning
ABox 1 A student who thinks any mixture of two substances is a solution picks this. A and B form separate layers, so the composition depends on where the sample is taken: this is a heterogeneous mixture.
BBox 2 A student who does not distinguish compounds from solutions picks this. Every particle here is the same A–B unit, held together by a bond, so the sample is a single compound, not a mixture.
CBox 3Correct This box contains two substances, A and B, as separate particles mixed evenly, so any part of the sample has the same composition as any other. That makes it a homogeneous mixture, a solution.
DBox 4 A student who thinks any liquid is a solution picks this. The box contains only particles of B, a single pure substance, so it is not a mixture of any kind.
A student dissolves 17.1 g of sucrose, C₁₂H₂₂O₁₁ (molar mass 342.3 g/mol), in 40.0 mL of water. The volume of the resulting solution is 50.5 mL. What is the molarity of sucrose in the solution?
Answer and reasoning
A0.989 MCorrect n = 17.1 g ÷ 342.3 g/mol = 0.0500 mol. Molarity uses the volume of the solution, 0.0505 L, not the volume of water: 0.0500 mol ÷ 0.0505 L = 0.989 M.
B1.25 M A student who divides by the volume of water, 0.0400 L, picks this. The dissolved sucrose adds about 10 mL to the volume, and molarity is moles per liter of solution, 0.0505 L.
C0.0500 M A student who treats the amount of sucrose as its concentration picks this. 0.0500 mol is the amount; the concentration is that amount divided by the volume of solution in liters.
D339 M A student who divides the mass of sucrose by the volume of solution picks this: 17.1 g ÷ 0.0505 L = 339 g/L. Molarity needs moles, so the mass must first be divided by the molar mass.
Working n = 17.1 g ÷ 342.3 g/mol = 0.04996 mol. The volume in M = nsolute/Lsolution is the volume of the solution, 50.5 mL = 0.0505 L. M = 0.04996 mol ÷ 0.0505 L = 0.989 M.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
3.7.A.1 Solution (homogeneous mixture) Fix
Solution (homogeneous mixture)
A mixture whose macroscopic properties, such as composition, density and color, do not vary throughout the sample. Solutions can be solids (for example an alloy such as brass), liquids (for example salt water) or gases (for example dry air).
Heterogeneous mixture
A mixture whose macroscopic properties depend on where in the mixture a sample is taken, for example a suspension of a solid in water whose density increases toward the bottom, or two liquids that form layers.
Solute and solvent
In a solution, the solvent is the component present in the larger amount (water in an aqueous solution) and a solute is a substance dispersed in it, as separate molecules or ions mixed evenly among the solvent particles.
Students often think Any mixture of two substances is a solution, even if the components form layers or the properties of the mixture vary from place to place. In fact No. A mixture is a solution only if it is homogeneous, with macroscopic properties that do not vary throughout. Mixtures whose components form layers, or whose properties depend on where the sample is taken, are heterogeneous.
Students often think Compounds and homogeneous mixtures are the same kind of matter, so a uniform sample made of two components, whether they are bonded together or simply mixed, is both a compound and a solution. In fact No. In a compound, atoms of different elements are bonded together in a fixed ratio, so every particle is the same and the composition is set by the formula. In a solution, two or more substances are mixed as separate particles; its composition can vary from sample to sample, and its components can be separated by physical means.
3.7.A.2 Molarity (M) Fix
Molarity (M)
The amount of solute in moles per liter of solution: M = nsolute/Lsolution. It is the most common way of expressing solution composition in the laboratory.
Volume of solution
The total volume of the solution after the solute has dissolved, which is the volume used in M = nsolute/Lsolution. It is generally not equal to the volume of solvent used, because the dissolved solute also changes the volume.
Number of solute particles
The amount of solute in a sample is M × V (in liters); multiplying by Avogadro's number, 6.022 × 10²³ mol⁻¹, gives the number of formula units or molecules. A dissolved ionic compound is present as separate ions, so the number of ions is found from the number of each ion in the formula, for example three ions (one Ca²⁺ and two Cl⁻) per CaCl₂.
Dilution
Adding solvent to a solution. The amount of solute stays the same while the volume of solution increases, so the molarity decreases: M₁V₁ = M₂V₂, where V₂ is the final volume of the solution.
Concentration of a portion
Because a solution has the same composition throughout, every portion taken from it has the same molarity as the whole; a smaller portion contains less solute in proportion to its smaller volume.
Volumetric flask
A flask with a single calibration mark on its neck, used to prepare a solution of accurately known volume: the solute is dissolved in some solvent in the flask, and solvent is then added until the bottom of the meniscus is on the mark.
Students often think The volume used to calculate molarity is the volume of solvent added, because the solute adds nothing to the volume. In fact No. It is the volume of the final solution. Dissolving a solute changes the volume, and in dilution the final volume includes the original solution as well as the added water.
Students often think The concentration of a solution is the same thing as the amount of solute it contains, so 0.100 M means 0.100 mol of solute in the sample. In fact No. Concentration is the amount of solute per unit volume of solution. A 0.100 M solution contains 0.100 mol of solute in each liter, so a 50.0 mL sample contains only 0.00500 mol.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
Two mixtures, X and Y, are each made by stirring a white solid into water. After each mixture has stood for one hour, samples are taken from its top, middle and bottom, and the density of each sample is measured. The table shows the results. Which claim is best supported by the data?
Answer and reasoning
AX is a compound of the solid and water, as X has one fixed density all the way through. A student who does not distinguish solutions from compounds picks this. A density that is the same throughout shows that X is homogeneous, a solution of the solid in water, not that a compound has formed.
BY is a solution in which the dissolved solid slowly settles toward the bottom over time. A student who thinks dissolved solute settles out of a solution picks this. The particles of a dissolved solute stay evenly mixed with the water; a mixture whose solid settles, so that its density varies with position, is heterogeneous.
CY is a heterogeneous mixture, because its density depends on where it is sampled.Correct The density of Y increases from 1.02 g/mL at the top to 1.21 g/mL at the bottom, so its macroscopic properties depend on location in the mixture. That is the defining property of a heterogeneous mixture; the solid in Y is suspended and settling, not dissolved.
DX and Y are both solutions, since each is a white solid stirred into water. A student who thinks any mixture is a solution picks this. Only X has the same density throughout; Y's density depends on location, so Y is a heterogeneous mixture.
A student dissolves sucrose (table sugar) in water, stirs until no solid remains, and leaves the clear solution to stand for a day. The student then draws the particulate diagram shown to represent the solution. Which statement best evaluates the diagram?
Answer and reasoning
AIt is consistent: once dissolved, the sucrose molecules gradually settle to the bottom. A student who thinks dissolved solute settles out of a solution picks this. Dissolved sucrose molecules stay evenly mixed with the water; only undissolved or suspended solid settles.
BIt is inconsistent: dissolved sucrose no longer exists, so it should not be drawn at all. A student who thinks a dissolved substance disappears picks this. The sucrose molecules are still present, spread among the water molecules; the sugar can be recovered by evaporating the water.
CIt is inconsistent: dissolved sucrose should be drawn as separate C, H and O atoms. A student who thinks dissolving breaks a molecular solute into atoms picks this. Dissolving separates whole sucrose molecules from one another; the covalent bonds inside each molecule do not break.
DIt is inconsistent: the sucrose molecules should be spread evenly among the water molecules.Correct Sucrose dissolved to form a solution, a homogeneous mixture. Its molecules keep moving randomly among the water molecules, so they stay evenly spread through the liquid however long it stands; a sample from the bottom has the same composition as one from the top.
What is the total number of solute particles from the dissolved CaCl₂ in 50.0 mL of 0.100 M CaCl₂(aq)?
Answer and reasoning
A3.01 × 10²¹ A student who thinks dissolved CaCl₂ stays as intact formula units counts one particle per formula unit: 0.00500 mol × 6.022 × 10²³ = 3.01 × 10²¹. In water the ions separate, giving three ions per formula unit.
B6.02 × 10²¹ A student who keeps the two chlorine atoms together as a Cl₂²⁻ ion counts two ions per formula unit. CaCl₂ contains Ca²⁺ and Cl⁻ ions in a 1:2 ratio, and each Cl⁻ ion separates on its own, giving three ions per formula unit.
C1.81 × 10²³ A student who takes 0.100 M to mean 0.100 mol of CaCl₂ in the sample picks this. The amount in 50.0 mL is 0.100 mol/L × 0.0500 L = 0.00500 mol.
D9.03 × 10²¹Correct The sample contains 0.100 mol/L × 0.0500 L = 0.00500 mol of CaCl₂. Each formula unit separates into one Ca²⁺ and two Cl⁻ ions, so there are 0.0150 mol of ions, or 0.0150 × 6.022 × 10²³ = 9.03 × 10²¹ ions.
Working n(CaCl₂) = 0.100 mol/L × 0.0500 L = 0.00500 mol. Each formula unit gives one Ca²⁺ and two Cl⁻ ions, three ions in all, so n(ions) = 3 × 0.00500 = 0.0150 mol. Number of ions = 0.0150 mol × 6.022 × 10²³ mol⁻¹ = 9.03 × 10²¹.
A bottle contains 500.0 mL of 0.400 M KNO₃(aq). A student pours 20.0 mL of the solution from the bottle into a beaker. What is the molarity of KNO₃ in the beaker?
Answer and reasoning
A0.0160 M A student who thinks a smaller portion is less concentrated scales the molarity by 20.0/500.0. The portion contains less solute, but in proportion to its smaller volume, so the molarity is unchanged.
B0.400 MCorrect A solution has the same composition throughout, so a portion has the same molarity as the whole. The 20.0 mL portion contains 0.00800 mol of KNO₃, and 0.00800 mol ÷ 0.0200 L = 0.400 M.
C10.0 M A student who applies M = n/V to the smaller volume while keeping the amount of solute fixed scales the molarity by 500.0/20.0. The amount of KNO₃ in the beaker is 25 times smaller as well, so M stays 0.400 M.
D0.00800 M A student who reports the amount of KNO₃ in the beaker, 0.00800 mol, as its concentration picks this. Molarity is that amount divided by the volume, 0.0200 L.
Working The solution is homogeneous, so the portion has the same composition as the whole. Check with M = n/V: n in the beaker = 0.400 mol/L × 0.0200 L = 0.00800 mol, and M = 0.00800 mol ÷ 0.0200 L = 0.400 M. Both n and V fall by the same factor (500.0/20.0 = 25), so their ratio is unchanged.
A student adds 600.0 mL of water to 200.0 mL of 0.500 M NaCl(aq), giving 800.0 mL of solution. What is the molarity of NaCl in the diluted solution?
Answer and reasoning
A0.125 MCorrect The diluted solution contains the same 0.100 mol of NaCl (0.500 mol/L × 0.2000 L) in 0.8000 L of solution, so M = 0.100 ÷ 0.8000 = 0.125 M.
B0.500 M A student who thinks adding water does not change the concentration, because no NaCl is added or removed, picks this. The same amount of NaCl is now spread through four times the volume, so the molarity is one-fourth as great.
C0.167 M A student who divides by the volume of water added, 0.6000 L, picks this. The final volume of the solution, 0.8000 L, includes the original 200.0 mL.
D0.100 M A student who reports the amount of NaCl, 0.100 mol, as the molarity picks this. The concentration is that amount divided by the volume of solution, 0.8000 L.
Working n(NaCl) = 0.500 mol/L × 0.2000 L = 0.100 mol, which is unchanged by dilution. M₂ = 0.100 mol ÷ 0.8000 L = 0.125 M (equivalently M₂ = M₁V₁/V₂ = 0.500 × 200.0/800.0).
A student prepares 100.0 mL of 0.200 M CuSO₄(aq) by placing 3.19 g of CuSO₄(s) in a 100.0 mL volumetric flask, dissolving it in some distilled water, adding distilled water to the mark and mixing. The molarity of the solution is later found to be less than 0.200 M. Which of the following could explain this result?
Answer and reasoning
AThe flask still held some distilled water from rinsing when the CuSO₄ was added. A student who thinks the water used, rather than the final volume, sets the molarity picks this. Because the flask is filled to the mark afterward, the final volume is still 100.0 mL and the molarity is unaffected.
BWater was added until the bottom of the meniscus was above the mark.Correct Filling past the mark makes the volume of solution greater than 100.0 mL while the amount of CuSO₄ stays at 0.0200 mol, so M = n/V is less than 0.200 M.
CHalf of the solution was poured into a beaker before its molarity was measured. A student who thinks a smaller portion of a solution is less concentrated picks this. The solution is homogeneous, so the half in the beaker has the same molarity as the whole.
DThe CuSO₄ dissolved completely, so no solid could be seen in the flask. A student who thinks a dissolved solute disappears picks this. Dissolving completely is what should happen; the Cu²⁺ and SO₄²⁻ ions are still present in the solution and the molarity is unaffected.
A student weighs a sample of KCl(s), dissolves all of it in water in a beaker, weighs the solution that forms, and wants to calculate the molarity of the KCl(aq). The mass of the KCl, the molar mass of KCl and the mass of the solution are known. Which additional quantity, if any, does the student need?
Answer and reasoning
AThe volume of the water that the KCl is dissolved in A student who thinks the volume in the molarity equation is the volume of solvent picks this. The dissolved KCl changes the volume, and M = nsolute/Lsolution uses the volume of the solution, which is not the volume of water that was added.
BNothing more, as grams of solution equal its milliliters A student who takes the mass of any aqueous solution in grams as its volume in milliliters picks this. That holds only for a liquid of density 1.00 g/mL; without the density of this KCl solution, its mass does not give its volume, so the volume still has to be measured.
CNothing more, as the moles of KCl are its molarity A student who treats the amount of solute as the concentration picks this. The moles of KCl are an amount; the molarity is that amount divided by the volume of the solution in liters, which has not been measured.
DThe volume of the solution that is formed in the beakerCorrect Molarity is moles of solute per liter of solution. The mass and molar mass give the moles of KCl, so the volume of the finished solution is the one quantity still needed.
Working M = nsolute/Lsolution. The moles of solute are fixed by the known quantities: n = mass of KCl ÷ molar mass of KCl. The quantity still missing is the denominator, the volume of the solution in liters. Neither the volume of the water used nor the mass of the solution is the volume of the solution: the density of KCl(aq) is not 1.00 g/mL.
The table gives the molarity and the volume of two solutions of KNO₃(aq). The two solutions are poured into one flask and mixed. Assuming that the volumes are additive, what is the molarity of KNO₃ in the mixture?
Answer and reasoning
A0.900 M A student who adds the two molarities picks this: 0.650 M + 0.250 M. The moles of KNO₃ add, but so do the volumes, and molarity is their ratio; a mixture cannot be more concentrated than both of its parts.
B0.450 M A student who takes the simple average of the two molarities picks this: (0.650 M + 0.250 M) ÷ 2. That would be right only for equal volumes; here three times as much of the 0.250 M solution is used, so the result lies closer to 0.250 M.
C0.140 M A student who treats the amount of solute as the concentration picks this: 0.0650 mol + 0.0750 mol = 0.140 mol. That is the total amount of KNO₃; the molarity is this amount divided by the total volume, 0.4000 L.
D0.350 MCorrect Total moles of KNO₃ = (0.650 M)(0.1000 L) + (0.250 M)(0.3000 L) = 0.0650 mol + 0.0750 mol = 0.1400 mol. Total volume = 0.4000 L, so M = 0.1400 mol ÷ 0.4000 L = 0.350 M.
Working n₁ = 0.650 mol/L × 0.1000 L = 0.0650 mol. n₂ = 0.250 mol/L × 0.3000 L = 0.0750 mol. Total n = 0.1400 mol. Total V = 0.1000 L + 0.3000 L = 0.4000 L. M = 0.1400 mol ÷ 0.4000 L = 0.350 M.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
