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 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

Practice 1 — answer: A

B calls water a universal solvent, but water dissolves only polar and ionic solutes; nonpolar grease is attracted to water far more weakly than water is to itself. C borrows "opposites attract" from charges, where it does not apply. Solubility follows like dissolves like, so the nonpolar grease needs a nonpolar solvent. D counts on scrubbing, which changes only the rate, not whether a mismatch ever 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

Practice 2 — answer: A

B leans on water as a universal solvent, but water dissolves only polar and ionic solutes; a nonpolar vitamin stays out, which is why the body stores it in fat rather than blood. C borrows "opposites attract" from charges, where solubility does not follow it. D counts on time, which changes only the rate, not whether a mismatch dissolves at all.

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

Worked example 1 — solution

Worked example 1 — solution

Worked example 1 — solution

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

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: 48 g (= 150 × 32/100) is what stays dissolved at 20 °C, not what fell out. B: 74 g forgot to scale — 106 − 32 is per 100 g of water, but 150 g is present. D: 207 g added the two amounts, 159 + 48, instead of subtracting.

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 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

Practice 1 — answer: B

A flipped the ratio: 40.0 × (0.400 ÷ 5.00) = 3.20 mL, a volume smaller than the start. C reported only the water added: 500. − 40.0 = 460. mL, dropping the original 40.0 mL. D stopped at the solute amount: 40.0 × 5.00 = 200 millimoles, 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

Practice 2 — answer: C

A flipped the ratio: 25.0 × (0.250 ÷ 2.00) = 3.13 mL, smaller than the start. B stopped at the solute amount: 25.0 × 2.00 = 50.0 millimoles, not a volume. D reported only the water added: 200. − 25.0 = 175. mL, dropping the original 25.0 mL.

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 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: C

A divided by grams of solvent: 0.402 ÷ 300 = 0.00134, in mol/g. B used the solution mass: 0.402 ÷ 0.330 kg = 1.22. D skipped the molar mass: 30.0 ÷ 0.300 = 100. Grams become moles first.

Worked example 3 — solution

A common first attempt

The unit came out mol/g. Molality is mol/kg — the solvent must enter in kilograms.

Worked example 3 — solution

The molar mass gives the moles: 45.0 g ÷ 60.06 g/mol = 0.749 mol urea.

Worked example 3 — solution

The 250. g of water is 0.250 kg. Divide:

Worked example 3 — solution

Practice 2 — answer: B

A divided by grams of solvent: 0.644 ÷ 500 = 0.00129, in mol/g. C used the solution mass: 0.644 ÷ 0.540 kg = 1.19. D skipped the molar mass: 40.0 ÷ 0.500 = 80.0. Convert grams to moles, then divide by kilograms of solvent.

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 — that is 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.

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: C

A 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 — solution

Step 2 · Molar mass → mass of solute

Two masses come out of that liter. The molarity fixes the moles, and the molar mass their mass: 4.50 mol × 98.08 g/mol = 441.4 g H₂SO₄.

Worked example 3 — solution

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: A

B skipped the density: 0.250 × 1000 ÷ 138.21 = 1.81, reading 1 L as 1000 g instead of 1240 g. C divided by the density: 0.250 × 1000 ÷ 1.24 ÷ 138.21 = 1.46. D dropped the 1000 mL: 0.250 × 1.24 ÷ 138.21 = 0.00224.

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 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

Practice 1 — answer: C

A forgot i: 1.86 × 0.90 = 1.67 °C uses i = 1, but KCl gives two ions. B used Kb: 2 × 0.512 × 0.90 = 0.92 °C is the boiling constant, not the freezing one. D dropped Kf: 2 × 0.90 = 1.80 leaves out 1.86 °C·kg/mol.

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

Worked example 3 — solution

Practice 2 — answer: C

A forgot i: 0 − (1.86 × 1.00) = −1.86 °C uses i = 1, but BaCl₂ gives three ions. B reported the depression: +5.58 °C skips the shift from 0. D dropped Kf: 3 × 1.00 = 3.00, so 0 − 3.00 = −3.00 °C leaves out 1.86.

Worked example 1 — solution

Step 1 · List what you know

5.0 mol of water is the solvent. 1.0 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 2 — solution

Step 1 · List what you know

6.0 mol benzene is the solvent. 2.0 mol nonvolatile solute. P° = 95.1 torr.

Worked example 2 — solution

The solute takes the rest: X_solute = 2.0 ÷ 8.0 = 0.250.

Worked example 2 — solution

Worked example 2 — the pressure lowering

The lowering ΔP

Worked example 2 — the pressure lowering

Practice 1 — answer: C

A used the solute's fraction: (1.0 ÷ 5.0) × 197.0 = 39.4 torr, the lowering not the pressure. B divided solute by solvent: (1.0 ÷ 4.0) × 197.0 = 49.25 torr, not a mole fraction. D assumed no change: 197.0 torr, but a nonvolatile solute lowers it.

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

Because P = X_solvent·P°, the pressure ratio is the SOLVENT's fraction, not the solute's:

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: B

A assumed no change: 43.9 torr, but a nonvolatile solute lowers it. C divided solute by solvent: (2.0 ÷ 6.0) × 43.9 = 14.6 torr, not a mole fraction. D used the solute's fraction: (2.0 ÷ 8.0) × 43.9 = 11.0 torr, the lowering not the pressure.

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 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

Practice 1 — answer: C

A inverted the proportion like Boyle's law: 0.043 × (1.0/5.0) = 0.0086 g/L, less dissolved though the pressure rose. B added the pressure change: 0.043 + (5.0 − 1.0) = 4.043 g/L, adding atm to g/L. D assumed no change: at 5.0 atm the solubility cannot stay 0.043 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

Practice 2 — answer: B

A inverted the proportion: 2.0 × (0.046/0.115) = 0.80 atm, less pressure for more gas. C added the change: 2.0 + (0.115 − 0.046) = 2.069 atm, adding g/L to atm. D assumed no change: keeping 2.0 atm leaves the solubility at 0.046 g/L, not 0.115.

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 titrant

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 2 — solution

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

Step 1 · Moles of titrant

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 by the analyte volume

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

Worked example 2 — the mole ratio

Practice 1 — answer: B

A swapped the two volumes: 0.300 × 25.00 ÷ 2 ÷ 30.00 = 0.125. C ignored the 2:1 ratio: 9.00 ÷ 25.00 = 0.360. D multiplied by the ratio: 2 × 9.00 ÷ 25.00 = 0.720.

Worked example 3 — solution

Three protons per acid make the ratio 3:1; otherwise the same three moves.

Step 1 · Moles of titrant

30.00 mL of 0.150 M NaOH delivers 4.50 mmol NaOH.

Worked example 3 — solution

Each H₃PO₄ takes three NaOH, so divide by three: 1.50 mmol H₃PO₄.

Worked example 3 — solution

Worked example 3 — solution

Practice 2 — answer: B

A swapped the two volumes: 0.200 × 20.00 ÷ 3 ÷ 45.00 = 0.0296. C ignored the 3:1 ratio: 9.00 ÷ 20.00 = 0.450. D multiplied by the ratio: 3 × 9.00 ÷ 20.00 = 1.35.