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 1 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.

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 2 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.

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

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 1 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 2 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.

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 1 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.

Worked example 3: 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 3: solution

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

Worked example 3: solution

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

Worked example 3: solution

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

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 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.

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

Worked example 2: solution

Grams → moles, each component

Ethylene glycol: 62.0 g ÷ 62.07 g/mol = 0.999 mol. Water: 180. g ÷ 18.02 g/mol = 9.99 mol.

Worked example 2: solution

Worked example 2: solution

Practice 1 answer: A

B divided by the water's moles alone: 0.499 ÷ 3.00 = 0.167. C skipped the molar masses: 23.0 ÷ 77.0 = 0.299 is a mass fraction. D stopped at 0.499 mol ethanol, before dividing by the total.

Worked example 3: 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 3: solution

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

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 2 answer: D

A kept the water in grams: 2.00 ÷ 1002 = 0.00200. B divided by the water alone: 2.00 ÷ 55.5 = 0.0360. C stopped at the water's moles. E is the water's share: 55.5 ÷ 57.5 = 0.965.

Worked example 1: solution

Three conversion factors carry 1 L of solution to moles of HCl.

Step 1 · Pick a basis

Take exactly 1 L (1000 mL) of the solution.

Worked example 1: solution

The density weighs that liter: 1000 mL × 1.19 g/mL = 1190 g of solution.

Worked example 1: solution

The label makes 37.0% of that mass HCl: 0.370 × 1190 g = 440.3 g HCl. The molar mass turns it into moles:

Worked example 1: solution

Worked example 1: solution

Worked example 2: solution

Step 1 · Pick a basis

Take 1 L (1000 mL) of the concentrated acid.

Worked example 2: solution

That treated 1 L as 1000 g. This acid is far denser than water.

Worked example 2: solution

At 1.84 g/mL the liter weighs 1000 mL × 1.84 = 1840 g, not 1000 g.

Worked example 2: solution

Worked example 2: solution

Practice 1 answer: A

C skipped the density: 0.150 × 1000 ÷ 40.00 = 3.75, reading 1 L as 1000 g instead of 1160 g. B divided by the density: 0.150 × 1000 ÷ 1.16 ÷ 40.00 = 3.23. D dropped the 1000 mL: 0.150 × 1.16 ÷ 40.00 = 0.00435, a thousand times too small.

Worked example 3: mass percent

Step 2 · Density links volume and mass

Worked example 3: mass percent

The molarity puts 4.50 mol in the liter, and the molar mass weighs them: 4.50 mol × 98.08 g/mol = 441.4 g H₂SO₄.

Worked example 3: mass percent

Worked example 3: mass percent

Worked example 3: molality

A common first attempt

That kilogram count includes the acid. Molality divides by the solvent alone.

Worked example 3: molality

The water is what remains: 1250 g − 441.4 g = 808.6 g = 0.809 kg.

Worked example 3: molality

Practice 2 answer: D

A skipped the density: 233.8 ÷ 1000 × 100 = 23.4. B divided by it: 1000 ÷ 1.15 = 870 g, 233.8 ÷ 870 × 100 = 26.9. C stopped at 234 g of NaCl, a mass, not a percent.

Guided example: solution

Two conversion factors are needed: the density and the molar mass.

Step 1 · Pick a basis

Take the batch as mixed. The solution is solute plus water: 18.0 g + 132 g = 150. g of solution.

Guided example: solution

Molarity needs liters. The density turns grams of solution into milliliters:

Guided example: solution

Guided example: solution

Practice 3 answer: E

A used the water's 150. mL as the syrup's volume: 0.2921 ÷ 0.150 = 1.95. D skipped the density: 0.2921 ÷ 0.250 = 1.17. B multiplied by it: 0.2921 ÷ 0.295 = 0.990. C left out the sucrose: 0.2921 ÷ 0.1271 = 2.30.

Guided example: the class

Step 1 · Run the beam test

The beam leaves a bright line. Particles large enough to scatter light are present, so the lotion is not a solution.

Guided example: the class

Nothing settles and the filter paper catches nothing. The droplets sit below suspension size.

Guided example: the class

Guided example: the type

Step 3 · Name the colloid type

Liquid in liquid is an emulsion, the same type as milk.

Guided example: the type

Practice 1 · answer: A

B swapped phase and medium: an aerosol is liquid or solid dispersed in a gas, like fog or a spray. C judged by the creamy look; an emulsion is liquid in liquid, like milk. D named the wrong dispersed phase; a sol is solid in liquid, like paint.

Practice 1 · answer: A

Practice 2 · answer: B

A swapped the two: a colloid never settles, and sample A drops a layer overnight. C sorted by the beam alone; both glow, but only B keeps its particles dispersed. D read B's clear filter paper and no settling as dissolved, but dissolved particles leave no beam track.

Practice 2 · answer: B

Practice 3 · answer: E

A sorted the ink by the beam alone; a suspension also settles and is caught, and A did neither. B took the ink, a liquid from the bottle, as the dispersed phase; the particles scattering the beam are solid carbon, and an emulsion is liquid in liquid, like milk. C sorted the saline by settling and the filter alone; solutions pass both too, and only the beam splits them. D read A's clear filter and no settling as dissolved; dissolved particles leave no beam track.

Practice 3 · answer: E

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: 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.

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

Practice 2 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.

Worked example 1: solution

Step 1 · List what you know

5.00 mol of water is the solvent. 1.00 mol of nonvolatile solute. P° = 23.8 torr. The unknown is P.

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 · List what you know

6.00 mol benzene is the solvent. 2.00 mol nonvolatile solute. P° = 95.1 torr.

Worked example 2: solution

The solute takes the rest: X_solute = 2.00 ÷ 8.00 = 0.250.

Worked example 2: solution

Worked example 2: the pressure lowering

The lowering ΔP

Worked example 2: the pressure lowering

Guided example: solution

Step 1 · List what you know

Raoult's law counts moles. Two conversion factors are needed, one molar mass for each substance.

Guided example: solution

Guided example: the vapor pressure

Step 2 · Solvent's mole fraction Step 3 · Apply Raoult's law

Guided example: the vapor pressure

Practice 1 answer: D

Both masses become moles first. A took a mass fraction: 90.0 ÷ 158.4 × 23.8 = 13.5 torr, but Raoult's law counts moles. B used the solute's fraction: 0.0385 × 23.8 = 0.916 torr, the lowering not the pressure. C left the sucrose in grams: 4.99 ÷ (4.99 + 68.4) × 23.8 = 1.62 torr.

Practice 1 answer: D

Worked example 3: solution

A common first attempt

32 mol of solute against 8.0 mol of benzene, yet the pressure barely fell.

Worked example 3: solution

8.0 mol benzene is the solvent. The solution reads P = 76.08 torr; pure benzene is P° = 95.1 torr. The unknown is the moles of nonvolatile solute.

Worked example 3: solution

Worked example 3: moles of solute

Step 3 · Apply Raoult's law

Solve X_solvent = mol benzene ÷ total for the total, then subtract the benzene:

Worked example 3: moles of solute

Practice 2 answer: A

The lowering gives the solute's share; the acetone holds the rest. B stopped at the total: 15.0 mol counts the acetone too. C read 0.200 as the solvent's share: 12.0 ÷ 0.200 = 60.0 total, then 60.0 − 12.0 = 48.0. D divided the acetone by the solute's share and stopped: 12.0 ÷ 0.200 = 60.0.

Practice 2 answer: A

Practice 3 answer: D

A stopped at the total, which counts the chloroform too. B kept the chloroform in grams: 215 ÷ 0.8000 − 215 = 53.8. C took the solute's 0.2000 share of the chloroform alone: 0.2000 × 1.801 = 0.360. E read 0.8000 as the solute's share: 1.801 ÷ 0.2000 − 1.801 = 7.20.

Practice 3 answer: D

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 1 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 2 answer: B

Practice 2 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.

Worked example 1: solution

Three moves: the titrant's millimoles, the one-to-one ratio, then divide by the acid's volume.

Step 1 · Moles of the known solution

The base's molarity is 0.200 mmol per mL. Its volume converts to millimoles; the mL cancel:

Worked example 1: solution

HCl and NaOH react one-to-one, so the acid supplied the same count: 5.00 mmol HCl.

Worked example 1: solution

The acid's millimoles over its volume give the molarity. Millimoles per milliliter is moles per liter:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Three moves, but the base-to-acid ratio is no longer one.

Step 1 · Moles of the known solution

The base delivers 0.250 mmol per mL, so 32.00 mL is 8.00 mmol NaOH.

Worked example 2: solution

Carry the base's millimoles straight to the acid's volume, as if the ratio were one:

Worked example 2: solution

Carry the base's millimoles straight to the acid's volume, as if the ratio were one:

Worked example 2: the mole ratio

Step 2 · Apply the mole ratio Step 3 · Divide to finish

The ratio, 1 H₂SO₄ per 2 NaOH, halves the count; the acid's volume then divides:

Worked example 2: the mole ratio

Worked example 2: the route on the map

Practice 1 answer: B

A stopped at 1.80 mmol of oxalic acid, before dividing by its 18.00 mL. C ignored the 2:1 ratio: 3.60 ÷ 18.00 = 0.200 M. D multiplied by the ratio: 2 × 3.60 ÷ 18.00 = 0.400 M.

Worked example 3: solution

This time the acid is the known solution, and the base's volume is the unknown.

Step 1 · Moles of the known solution

25.00 mL of 0.0600 M H₃PO₄ holds 25.00 × 0.0600 = 1.50 mmol H₃PO₄.

Worked example 3: solution

Worked example 3: the mole ratio

Step 2 · Apply the mole ratio Step 3 · Divide to finish

The ratio, 3 NaOH per H₃PO₄, triples the count; the base's molarity, upside down, turns millimoles into milliliters:

Worked example 3: the mole ratio

Worked example 3: the route on the map

Practice 2 answer: A

Practice 2 answer: A

B ignored the 3:1 ratio: 3.168 mmol → 0.609 g. C stopped at 1.06 mmol, before the molar mass. D skipped mmol → mol: 1.056 × 192.12 = 203 g. E multiplied by the ratio: 9.504 mmol → 1.83 g.