Guided example: solution

Step 1 · Grams → moles

The amount is already in moles: 0.300 mol glucose. No molar mass is needed.

Guided example: solution

The volume is already in liters: 1.50 L of solution.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer D

A divided by the water added: 0.270 ÷ 1.15 = 0.235, but the dissolved sugar takes up room too. B flipped the fraction: 1.20 ÷ 0.270 = 4.44, in L/mol. C multiplied: 0.270 × 1.20 = 0.324, in mol·L.

Worked example 1: solution

Step 1 · Grams → moles

The mole count is given directly: 0.350 mol NaCl, no mass conversion needed.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Grams → moles

The molar mass converts the mass to moles: 9.35 g ÷ 74.55 g/mol = 0.1254 mol KCl.

Worked example 2: solution

The unit came out mol/mL. Molarity is mol/L: the volume must enter in liters.

Worked example 2: solution

The mark reads 250.0 mL of solution, which is 0.2500 L. Divide:

Worked example 2: solution

Worked example 2: the route on the map

Practice 2: answer B

A stopped at moles: 34.8 ÷ 101.11 = 0.344 mol is the amount in the sample. C divided by milliliters: 0.344 ÷ 225 = 0.00153, in mol/mL. D found the percent: 34.8 ÷ 225 × 100 = 15.5% (m/v). E divided grams by liters: 34.8 ÷ 0.225 = 155 g/L.

Practice 3: answer A

B divided by the whole batch: 0.0200 ÷ 2.000 = 0.0100, but the 2.38 g came from only 25.00 mL. C stopped at moles: 0.0200 mol is the amount in the portion. D divided by milliliters: 0.0200 ÷ 25.00 = 0.000800, in mol/mL. E multiplied by the volume instead of dividing: 0.0200 × 0.02500 = 0.000500.

Worked example 3: solution

Two conversion factors are needed.

Step 1 · Grams → moles

Worked example 3: solution

Two orientations exist. Only one cancels mol NaCl:

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

Three conversion factors are needed.

Step 1 · Grams → moles

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 4: answer D

A stopped at liters and relabeled: 10.1 ÷ 101.11 ÷ 0.400 = 0.250 L, which is 250. mL, not 0.250 mL. B flipped the molarity fraction: 10.1 ÷ 101.11 × 0.400 × 1000 = 40.0, and the units, mol²/L, are not a volume. C stopped at moles: 10.1 ÷ 101.11 = 0.0999 mol is the KNO₃ needed, not the volume that carries it.

Worked example 5: solution

Two conversion factors are needed.

Liters → moles: the molarity

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the map

Worked example 5: making the solution

Weigh 4.0 g of NaOH. Dissolve it in part of the water in a 1.0 L volumetric flask. Add water to the mark, then mix.

Worked example 5: making the solution

Weigh 4.0 g of NaOH. Dissolve it in part of the water in a 1.0 L volumetric flask. Add water to the mark, then mix.

Practice 5: answer B

A left out the volume: 0.150 × 142.04 = 21.3 g is the mass in a full liter. C flipped the molarity: 0.5000 ÷ 0.150 × 142.04 = 473, and the units do not cancel. D skipped mL → L: 500.0 × 0.150 × 142.04 = 1.07 × 10⁴.

Worked example 6: solution

Two conversion factors are needed for the mass.

Liters → moles: the molarity

Worked example 6: solution

Worked example 6: solution

Worked example 6: the percent

Grams per 100 mL: the % (m/v)

Keep the unrounded 20.6 g in the calculator. The blood is the solution: 5.6 L = 5.6 × 10³ mL.

Worked example 6: the percent

Grams per 100 mL: the % (m/v)

Keep the unrounded 20.6 g in the calculator. The blood is the solution: 5.6 L = 5.6 × 10³ mL.

Worked example 6: the percent

Grams per 100 mL: the % (m/v)

Keep the unrounded 20.6 g in the calculator. The blood is the solution: 5.6 L = 5.6 × 10³ mL.

Worked example 6: the route on the map

Practice 6: answer C

A filled one flask: 0.2500 × 0.200 × 159.61 = 7.98. B flipped the molarity: 1.500 ÷ 0.200 × 159.61 = 1.20 × 10³. D stopped at moles: 1.500 × 0.200 = 0.300 mol. E skipped mL → L: 1500. × 0.200 × 159.61 = 4.79 × 10⁴.

Practice 6: the route on the map

Guided example: solution

Step 1 · List the knowns

Before, the stock: M₁ = 10.0 M, and V₁ is the unknown. After, the flask: M₂ = 0.200 M and V₂ = 250. mL.

Guided example: solution

Guided example: solution

Both concentrations are in M, so they cancel. V₂ is in mL, so V₁ comes out in mL.

Guided example: solution

At the bench: measure 5.00 mL of the stock, then add water to the 250. mL mark.

Guided example: solution

Guided example: the route on the map

Practice 1 answer: D

A flipped the ratio: 300. × (12.0 ÷ 0.600) = 6.00 × 10³ mL, twenty times the batch. B stopped at the solute: 300. × 0.600 = 180. mmol of HCl, not a volume. C reported the water: 300. − 15.0 = 285 mL goes in after the stock.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · List the knowns

M₁ = 6.00 M. V₁ = 25.0 mL. V₂ = 150. mL. The unknown is M₂.

Worked example 1: solution

Worked example 1: solution

Both volumes are in mL, so the volume ratio cancels to a pure number.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

A dilution that ends with less liquid than it started. The concentration ratio is upside down.

Worked example 2: solution

M₁ = 6.00 M. V₁ = 30.0 mL. M₂ = 0.500 M. The unknown is V₂.

Worked example 2: solution

Solve M₁V₁ = M₂V₂ for V₂: it equals V₁ scaled by the concentration ratio M₁ ÷ M₂.

Worked example 2: solution

V₁ is in mL, so V₂ comes out in mL. The starting concentration sits on top:

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 answer: B

A stopped at the total volume: 500. mL still counts the 40.0 mL of acid already in the flask. C flipped the ratio: 40.0 × (0.400 ÷ 5.00) = 3.20 mL, less than the start. D stopped at the solute: 40.0 × 5.00 = 200. mmol of HNO₃, not a volume.

Practice 2: the route on the map

Worked example 3: solution

A common first attempt

50.0 mL is only the water added. A percent counts per 100 mL of solution, and the solution is all 60.0 mL: stock plus water.

Worked example 3: solution

C₁ = 9.00% (m/v). V₁ = 10.0 mL. V₂ = 60.0 mL, the whole solution. The unknown is C₂.

Worked example 3: solution

C₂ = C₁ × V₁ ÷ V₂. Both volumes are in mL, so C₂ comes out in % (m/v), the unit of C₁.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3 answer: A

B divided by the water alone, per solvent: 8.00 × (15.0 ÷ 25.0) = 4.80. C flipped the volume ratio: 8.00 × (40.0 ÷ 15.0) = 21.3, stronger than the start. D stopped at the solute: 15.0 mL × 8.00 g/100 mL = 1.20 g of KCl, not a percent.

Practice 3: the route on the map

Worked example 4: solution

A common first attempt

A droplet cannot dilute to fill 2.00 L. The result came out in liters, because the volume entered in liters.

Worked example 4: solution

M₂ = 0.150 M. V₂ = 2.00 L. M₁ = 6.00 M. The unknown is V₁, wanted in mL.

Worked example 4: solution

Solve M₁V₁ = M₂V₂ for V₁: it equals V₂ scaled by the concentration ratio M₂ ÷ M₁.

Worked example 4: solution

V₂ entered in liters, so V₁ lands in liters. Convert to milliliters at the end:

Worked example 4: solution

Worked example 4: the route on the map

Practice 4 answer: C

A stopped at the stock: 80.0 mL is what gets measured, and water fills the rest. B stopped at the solute: 0.900 × 0.400 = 0.360 mol of KOH, not a volume. D subtracted in liters: 0.900 − 0.0800 = 0.820 L, not mL; convert both volumes first.

Practice 4: the route on the map

Practice 5 answer: E

A stopped at the stock: 150. mL is what gets measured. B read 0.900% per solvent, 27.0 g NaCl per 3.00 × 10³ mL of water; the stock fills part of the 3.00 L. C subtracted in liters: 3.00 − 0.150 = 2.85 L, not mL. D stopped at the solute: 3.00 × 10³ mL × 0.900 g/100 mL = 27.0 g of NaCl.

Practice 5: the route on the map

Guided example: solution

Step 1 · Find the solute's polarity

Each C–Cl bond is polar, but the four identical arrows point to the corners of a tetrahedron and cancel.

Guided example: solution

Hexane carries no charges and no polar groups.

Guided example: solution

Broken: CCl₄–CCl₄ and hexane–hexane dispersion. Formed: CCl₄–hexane dispersion.

Guided example: solution

Guided example: the route on the map

Practice 1: answer B

A paired an ionic solid with a nonpolar liquid, as if opposite polarities attract. C read CCl₄'s polar C–Cl bonds as a polar molecule; its symmetric shape cancels them. D treated water as the only polar solvent; methanol's O–H group makes it polar too.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Find the solute's polarity

Salt is built from Na⁺ and Cl⁻ ions. Their charge attracts polar molecules strongly, so ionic solutes belong with the polar family.

Worked example 1: solution

Water is polar. Each molecule has a positive end and a negative end that can turn toward a charge.

Worked example 1: solution

Water molecules surround each ion, their charged ends replacing the pull the ions had on one another.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first answer

Water dissolves a lot, but only polar and ionic solutes. Adding more water changes nothing here.

Worked example 2: solution

Grease is a nonpolar mix of oils. Its molecules attract each other only weakly and carry no charge.

Worked example 2: solution

Water is polar, and holds itself together with strong hydrogen bonds.

Worked example 2: solution

Water's hydrogen bonds to itself are far stronger than its weak attraction to the grease, so the grease is left untouched.

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 answer: C

A treats ionic solids as dissolving anywhere; ions need a polar solvent whose charged ends can surround them, and hexane has none. B judged glycerol by its carbon atoms; its three –OH groups make it polar, so it mixes with water, not hexane. D counts on stirring, which changes only the rate, never whether a mismatch dissolves.

Practice 2: the route on the map

Worked example 3: the polar groups

Step 1 · Find the solute's polarity

The carbon framework is nonpolar, but the molecule carries eight polar –OH groups.

Worked example 3: the polar groups

Step 1 · Find the solute's polarity

The carbon framework is nonpolar, but the molecule carries eight polar –OH groups.

Worked example 3: the polar groups

Step 1 · Find the solute's polarity

The carbon framework is nonpolar, but the molecule carries eight polar –OH groups.

Water is polar, and every one of those –OH groups can hydrogen-bond to it.

Worked example 3: the match

Step 3 · Match them

Worked example 3: the match

Step 3 · Match them

Worked example 3: the route on the map

Practice 3: answer A

B counted any –OH as enough; one –OH cannot carry a 20-carbon chain into water. C swapped the two families. D judged both by the carbon framework; vitamin C's four –OH groups make it polar.

Practice 3: the route on the map

Practice 4: answer E

A reversed bond breaking: pulling ions apart always costs energy. B reversed bond forming: forming ion–dipole forces releases energy. C read the cooling backwards: dissolving draws heat from the water, so the process absorbs it. D kept the label but flipped the comparison; a larger payback would warm the solution.

Practice 5 answer: D

CuSO₄ is ionic, so it belongs with polar water. Erythritol is built on carbon, but an –OH on every carbon makes it polar: water again. β-Carotene holds only C and H, so it is nonpolar: hexane.

A treated water as a universal solvent; it leaves nonpolar β-carotene out. B judged erythritol by its carbon skeleton instead of its –OH groups. C swapped the families, as if opposite polarities attract.

Practice 5: the route on the map

Guided example: solution

Step 1 · Read the curve at the temperature

The KNO₃ curve at 30 °C reads 46 g per 100 g of water.

Guided example: solution

The beaker holds exactly 100 g of water, so the reading is the limit as it stands.

Guided example: solution

46 g dissolves, then dissolving stops. The liquid holds the maximum, so it is saturated, not supersaturated.

Guided example: solution

Guided example: the route on the map

Worked example 1: solution

A common first attempt

A first attempt uses 64 g as the limit. But 64 g is per 100 g of water, and this beaker holds 200 g. Scale it first.

Worked example 1: solution

At 40 °C the KNO₃ curve reads 64 g per 100 g of water. Scale that to the 200 g in this beaker:

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Practice 1 answer: A

B scaled the wrong way: 34.0 × (100/450.) = 7.56. C used 34.0 g unscaled, the limit for 100 mL of water. D flipped the solubility: 450. × (100/34.0) = 1.32 × 10³, and mL H₂O does not cancel.

Practice 1: the route on the map

Practice 2 answer: B

A stopped at the limit: 180. g counts the 75.0 g already in. C added instead of subtracting: 180. + 75.0 = 255. D used 144 g unscaled: 144 − 75.0 = 69.0.

Practice 2: the route on the map

Practice 3 answer: C

A called beaker 3 supersaturated, but its extra stays solid. B compared every beaker with 0.15 g unscaled. D called beaker 1 unsaturated, yet 0.30 g is its full limit.

Practice 3: the route on the map

Worked example 2: solution

Step 1 · Read the curve at the temperature

Worked example 2: solution

The beaker holds 100 g of water, so the per-100-g values apply directly: 106 g dissolved at 60 °C, 32 g the limit at 20 °C.

Worked example 2: solution

At 60 °C all 106 g were dissolved. At 20 °C only 32 g can stay. The rest leaves solution as solid.

Worked example 2: solution

Worked example 2: the route on the map

Practice 4 answer: D

A: 0 g assumes stirring dissolves everything; past the limit the extra stays solid. B: 18 g is how much dissolves, not what is left: 75 − 18 = 57. C: 39 g used 36 g as the limit for this beaker, but 36 g is per 100 g of water: only 50. g is here, so 50. × 36/100 = 18 g dissolves.

Practice 4: the route on the map

Practice 5 answer: A

B stopped at 4.80 g, what stays dissolved. C used both readings unscaled: 73.0 − 12.0 = 61.0. D scaled the 80 °C reading but not the 20 °C one: 29.2 − 12.0 = 17.2.

Practice 5: the route on the map

Practice 6 answer: C

A assumed a saturated start: 340. − 115 = 225 g, but only 180. g was dissolved. B stopped at 115 g, what stays dissolved. D used 46.0 g unscaled: 180. − 46.0 = 134 g.

Practice 6: the route on the map

Guided example: solution

The O₂ pressure halves, so the dissolved O₂ should halve too: about 4.5 mg/L. The four method steps give the exact value.

Guided example: solution

Given: S₁ = 9.0 mg/L at P₁ = 160 mmHg, and P₂ = 80. mmHg. Same gas, same temperature. Wanted: S₂, the new solubility.

Guided example: solution

One gas at one temperature, so S/P stays constant: S₁/P₁ = S₂/P₂.

Guided example: solution

Multiply both sides by P₂. The new pressure lands on top.

Guided example: solution

Guided example: the route on the strip

Practice 1 answer: D

A inverted the proportion like Boyle's law: 42.0 × (2.40/0.800) = 126 mg/L, more N₂ held at lower pressure. B scaled by the pressure change: 42.0 × (2.40 − 0.800)/2.40 = 28.0 mg/L, the N₂ that comes out, not what stays. C assumed no change: at a third of the pressure, 42.0 mg/L cannot stay dissolved.

Worked example 1: solution

Step 1 · Identify given and wanted

S₁ = 1.45 g/L at P₁ = 1.0 atm. P₂ = 4.0 atm. The gas and the temperature do not change. The unknown is S₂.

Worked example 1: solution

Same gas, constant temperature, so S/P stays constant: S₁/P₁ = S₂/P₂.

Worked example 1: solution

The new pressure goes on top: solubility climbs with pressure.

Worked example 1: solution

Worked example 1: the route on the strip

Worked example 2: solution

Step 1 · Identify given and wanted

S₁ = 5.80 g/L at P₁ = 4.0 atm. P₂ = 1.0 atm. The gas and the temperature are unchanged. The unknown is S₂.

Worked example 2: solution

Same gas, constant temperature, so S/P stays constant: S₁/P₁ = S₂/P₂.

Worked example 2: solution

The new pressure still goes on top; here it is the smaller one.

Worked example 2: solution

Worked example 2: the route on the strip

Practice 2 answer: C

A inverted the proportion like Boyle's law: 0.0430 × (1.00/3.00) = 0.0143 g/L, less dissolved though the pressure rose. B scaled by the pressure change: 0.0430 × (3.00 − 1.00)/1.00 = 0.0860 g/L; the ratio takes the new pressure, 3.00 atm, not the 2.00 atm increase. D assumed no change: at 3.00 atm the solubility cannot stay 0.0430 g/L.

Worked example 3: solution

A common first attempt

Less pressure to dissolve more gas is impossible. Inverting the ratio follows Boyle's law, not Henry's law.

Worked example 3: solution

S₁ = 1.45 g/L at P₁ = 1.0 atm. The target is S₂ = 8.70 g/L. The unknown is P₂.

Worked example 3: solution

Same gas, constant temperature, so S/P stays constant: S₁/P₁ = S₂/P₂.

Worked example 3: the pressure needed

Step 3 · Rearrange for the unknown

The larger solubility goes on top, so the pressure comes out larger.

Worked example 3: the pressure needed

Worked example 3: the route on the strip

Practice 3 answer: A

B inverted the ratio like Boyle's law: 4.20 × (0.140/0.0850) = 6.92 atm, more pressure for less dissolved gas. C stopped at the ratio: 0.0850/0.140 = 0.607, never multiplied by P₁. D assumed no change: a lower concentration needs a lower pressure.

Practice 4 answer: B

Practice 4 answer: B

A is all the CH₄ dissolved at 6.00 atm: 0.138 × 0.400 = 0.0552 g. C skipped the ÷ 2.00 atm: 0.0828 g. D left mL unconverted: 41.4 g.

Practice 4: the route on the strip

Guided example: solution

Step 2 · Match the units

m/v pairs grams of solute with milliliters of solution. The whole converts to milliliters:

Guided example: solution

Part over whole, × 100, with the whole in milliliters. Left in liters, the fraction runs far above 100:

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer C

A compared the ethanol to the other 140. mL alone: 260. / 140. × 100 = 186. B stopped at the decimal fraction: 260. / 400. = 0.650. D flipped part and whole: 400. / 260. × 100 = 154. E added the ethanol to a volume that already held it: 260. / 660. × 100 = 39.4.

Worked example 1: solution

A common first attempt

That ratio compares the sucrose to the water alone. A mass percent compares it to the whole solution.

Worked example 1: solution

The whole is everything in the beaker, solute plus solvent: 25.0 g + 100.0 g = 125.0 g of solution.

Worked example 1: solution

Both measurements are masses: m/m, grams over grams. Part over whole, × 100.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

One conversion factor is needed.

Step 2 · Match the units

m/m pairs grams of solute with grams of solution, so the label declares 3.0 g H₂O₂ = 100 g solution.

Worked example 2: solution

The equality gives two orientations. Only one cancels the given unit, g solution:

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2: answer A

B divided by the water alone: 20.0 / 230.0 × 100 = 8.70. C flipped part and whole: 250.0 / 20.0 × 100 = 1.25 × 10³, and no percent concentration tops 100. D stopped at the decimal fraction: 20.0 / 250.0 = 0.0800; the × 100 makes it 8.00 per hundred.

Worked example 3: solution

One conversion factor is needed.

Step 2 · Match the units

m/v pairs grams of solute with milliliters of solution: the one percent that crosses between mass and volume, the way a density does.

Worked example 3: solution

The given is a mass of NaCl, so g NaCl belongs in the denominator: 100 mL solution over 0.90 g NaCl.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 3: answer D

Two conversion factors are needed.

A read the label as 15.0 g per 100 mL of water, as if 115 parts made the solution: 500. × 15.0 / 115 = 65.2. B skipped liters to milliliters: 0.500 × 15.0 / 100 = 0.0750. C flipped the label factor: 500. × 100 / 15.0 = 3.33 × 10³. E used the raw percent: 500. × 15.0 = 7.50 × 10³.

Practice 3: the route on the map

Practice 4: answer B

A skipped the density: 250. × 28.0 / 100 = 70.0 treats mL of solution as grams. C stopped halfway: 250. × 0.900 = 225 g is the whole solution, not the NH₃ in it. D flipped the density: 250. ÷ 0.900 = 278 g, then × 28.0 / 100 = 77.8.

Practice 4: the route on the map

Guided example: solution

Per kilogram: 800. g is less than a kilogram of water, so a full kilogram would carry more than 1.20 mol.

Guided example: solution

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1 answer: C

A multiplied instead of dividing: 0.840 × 0.600 = 0.504. B divided by grams of water: 0.840 ÷ 600 = 1.40 × 10⁻³, 1000 times too small. D flipped the fraction: 0.600 ÷ 0.840 = 0.714 counts kilograms per mole.

Worked example 1: solution

Step 1 · Grams → moles

The molar mass converts the solute's mass to moles: 18.0 g ÷ 180.16 g/mol = 0.0999 mol glucose.

Worked example 1: solution

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Practice 2 answer: A

B used the solution's mass: 0.500 mol ÷ 0.446 kg = 1.12. C skipped the molar mass: 46.0 g ÷ 0.400 kg = 115. D divided by grams of water: 0.500 ÷ 400 = 0.00125 mol/g, 1000 times too small.

Worked example 2: solution

A common first attempt

Forty-five kilograms of urea in a cup of water. The grams of water never cancel the kilograms in the molality.

Worked example 2: solution

The molality counts per kilogram of solvent: 250. g water = 0.250 kg water.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 3 answer: B

A stopped at moles: 0.750 × 1.60 = 1.20 mol. C divided the kilograms by the molality: 0.750 ÷ 1.60 = 0.469 mol, then × 76.09 = 35.7. D kept the water in grams: 750 × 1.60 × 76.09 = 9.13 × 10⁴.

Practice 4 answer: D

A flipped the molality factor: 0.200 × 0.500 = 0.100 kg, or 100. g. B skipped the molar mass: 8.00 ÷ 0.500 = 16.0 kg, or 1.60 × 10⁴ g. C stopped at 0.400 kg.

Practice 4: the route on the map

Practice 5 answer: E

A used the solution's mass: 0.500 ÷ 0.2873 = 1.74. B stopped at moles. C added the KCl to a total that already held it: 0.500 ÷ 0.3246 = 1.54. D divided by grams: 0.500 ÷ 250.0 = 2.00 × 10⁻³.

Practice 5: the route on the map

Guided example: solution

Step 1 · Count the particles

Guided example: solution

Same water, same Kb. KBr puts twice the particles in the water, so its boiling point rises twice as far.

Guided example: solution

Guided example: the route on the map

Worked example 1: solution

Step 1 · Count the particles

Ethylene glycol is molecular. It dissolves as whole molecules, so i = 1.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

NaCl is not one particle. This left out i.

Worked example 2: solution

NaCl dissolves into Na⁺ and Cl⁻: two particles per formula unit, so i = 2.

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: B

A judged by the size of the molecule: sucrose's 45 atoms stay together as one particle, so i = 1. C split the nitrate into atoms, 1 + 1 + 3 = 5 pieces; a polyatomic ion stays whole, so KNO₃ gives K⁺ and NO₃⁻, i = 2. D ignored i: equal molality is not equal particle count; 0.40 m of CaCl₂ is 1.20 mol/kg of particles.

Practice 1: the route on the map

Practice 2 answer: C

A forgot i: 1.86 × 0.900 = 1.67 °C uses i = 1, but KCl gives two ions. B used Kb: 2 × 0.512 × 0.900 = 0.922 °C is the boiling constant, not the freezing one. D stopped at i × m: 2 × 0.900 = 1.80 mol/kg counts the dissolved particles but never multiplies by Kf.

Practice 2: the route on the map

Worked example 3: solution

Step 1 · Count the particles

CaCl₂ dissolves into one Ca²⁺ and two Cl⁻: three particles, so i = 3.

Worked example 3: solution

The denominator is the water alone, 0.500 kg, not the 0.5416 kg of solution.

Worked example 3: the new freezing point

Step 3 · Multiply i · K · m

Worked example 3: the new freezing point

Worked example 3: the route on the map

Your turn: the route on the map

Practice 3 answer: A

B counted one NH₄⁺ and one SO₄²⁻: i = 2 gives −4.84 °C, but the 2 outside the parentheses means two ammonium ions. C forgot i: 0 − 1.86 × 1.30 = −2.42 °C. D stopped at the depression and dropped the sign: +7.25 °C is above 0 °C. E used Kb: 0 − 3 × 0.512 × 1.30 = −2.00 °C.

Practice 3: the route on the map

Practice 4 answer: D

A forgot i: 0 − 1.86 × 0.699 = −1.30 °C, but BaCl₂ gives three ions. B stopped at the depression and dropped the sign: +3.90 °C is above 0 °C. C divided by the solution's mass: 0.1748 ÷ 0.2864 kg = 0.610 m, giving −3.41 °C.

Practice 4: the route on the map

Worked example 4: solution

Step 1 · Count the particles

NaCl dissolves into Na⁺ and Cl⁻, so i = 2.

Worked example 4: solution

Worked example 4: solution

Worked example 4: grams of salt

Molality to grams

Two conversion factors are needed.

Worked example 4: grams of salt

Worked example 4: the route on the map

Practice 5 answer: C

A stopped at moles: 0.914 mol is one step short of grams. B forgot i: 8.50 ÷ 1.86 × 0.600 × 95.21 = 261 g. D flipped the rearrangement: (3 × 1.86) ÷ 8.50 × 0.600 × 95.21 = 37.5 g. E used Kb: 8.50 ÷ (3 × 0.512) × 0.600 × 95.21 = 316 g.

Practice 5: the route on the map

Guided example: solution

Step 1 · Mass fraction, same units

Milligrams to grams, then solute over solution.

Guided example: solution

One fraction, three powers of ten.

Guided example: solution

Guided example: the route on the map

Worked example: solution

Step 1 · Mass fraction, same units

Worked example: solution

Step 1 · Mass fraction, same units

Worked example: solution

Step 1 · Mass fraction, same units

Worked example: the route on the map

Practice 1: answer C

A used 10⁹: 1.8 × 10⁻⁴ ÷ 200.0 × 10⁹ = 900 is ppb. B stopped at the mass fraction: 1.8 × 10⁻⁴ ÷ 200.0 = 9.0 × 10⁻⁷. D used × 100: 9.0 × 10⁻⁵ is the mass percent.

Practice 1: the route on the map

Your turn: the route on the map

Practice 2: answer A

B kept milligrams over grams: 2.40 L is about 2400 g, and 0.84 ÷ 2400 × 10⁶ = 350, a thousand times too high. C flipped the fraction: 2.40 ÷ 0.84 = 2.9. D multiplied: 0.84 × 2.40 = 2.0.

Practice 2: the route on the map

Practice 3: answer E

Two conversion factors are needed.

A kept milligrams over grams: 150. mL is about 150. g, and 0.036 ÷ 150. × 10⁶ = 240. B multiplied: 0.036 × 0.150 = 0.0054. C flipped: 0.150 ÷ 0.036 = 4.2. D skipped milliliters to liters: 0.036 ÷ 150. = 2.4 × 10⁻⁴.

Practice 3: the route on the map

Practice 4: answer D

A used 10⁶: 0.016 is the same lead in ppm. B stopped at the mass fraction: 2.4 × 10⁻⁵ ÷ 1500 = 1.6 × 10⁻⁸. C used × 100: 1.6 × 10⁻⁶ is the mass percent.

Practice 4: the route on the map

Practice 5: answer B

A stopped at the mass fraction: 0.00416 ÷ 208.0 = 2.00 × 10⁻⁵, never scaled by 10⁶. C used the water shortcut: 4.16 mg ÷ 0.1600 L = 26.0 mg/L, read as ppm; that needs 1 L ≈ 1 kg. D flipped the density: 160.0 ÷ 1.30 = 123 g, and 0.00416 ÷ 123 × 10⁶ = 33.8.

Practice 5: the route on the map

Guided example: solution

Two conversion factors are needed.

Step 1 · Volume → moles

No mL → L hop: the volume is already 0.500 L. The molarity, 0.100 mol per liter, converts it to 0.0500 mol NaCl.

Guided example: solution

The mole ratio, written AgCl over NaCl (1 : 1), crosses substances; mol NaCl cancels:

Guided example: solution

Moles of AgCl were wanted. No third factor is needed.

Guided example: solution

Guided example: the route on the map

Practice 1: answer B

A read NaCl's coefficient for CaCO₃: 0.0100 × 2 = 0.0200. C inverted the molarity: 0.0400 ÷ 0.250 = 0.160. D fed milliliters to the molarity: 40.0 × 0.250 = 10.0.

Practice 1: the route on the map

Worked example 1: solution

Two conversion factors are needed.

Step 1 · Volume → moles

250.0 mL is 0.2500 L. The label's molarity converts the liters to moles of HCl:

Worked example 1: solution

The mole ratio, written H₂ over HCl (1 : 2), crosses substances; mol HCl cancels:

Worked example 1: solution

Worked example 1: the route on the map

Practice 2: answer C

A skipped the mole ratio: 0.0600 × 0.150 = 0.00900 mol is the AgNO₃. B inverted the ratio: 0.00900 × 3 = 0.0270. D fed milliliters to the molarity: 60.0 × 0.150 ÷ 3 = 3.00.

Practice 2: the route on the map

Worked example 2: solution

Three conversion factors are needed.

Step 1 · Volume → moles

150.0 mL is 0.1500 L. The molarity, 0.300 mol per liter, converts the liters to 0.0450 mol KI.

Worked example 2: solution

mol KI cannot cancel mol PbI₂. The equation makes one PbI₂ from two KI; the mole ratio must cross first.

Worked example 2: solution

The ratio, written PbI₂ over KI (1 : 2), crosses substances; the molar mass then converts out:

Worked example 2: solution

Worked example 2: the route on the map