AP Chemistry · Lesson 3

Solutions, molarity and separations

Know Know it

Molarity is moles of solute per liter of solution: M=nVM = \frac{n}{V}. Convert mL to L and grams to moles before dividing.

In a dilution, solvent is added and the moles of solute stay the same, so M1V1=M2V2M_1 V_1 = M_2 V_2. The volumes only need the same unit on both sides.

A soluble ionic compound separates into ions. A 0.10 M solution of CaCl₂ is 0.10 M in Ca²⁺ and 0.20 M in Cl⁻, because each formula unit gives two chloride ions. When two solutions are mixed, find the moles of each ion, add them, and divide by the total volume.

Like dissolves like: a polar solute dissolves in a polar solvent and a nonpolar solute in a nonpolar solvent, because the solute-solvent attractions must replace the attractions in the pure substances.

In paper or thin-layer chromatography, the stationary phase (paper or silica) is polar. A spot that is attracted more strongly to the stationary phase moves less, so it has a smaller RfR_f. In distillation, the component with weaker intermolecular forces boils first.

Worked example Easy

A student dissolves 4.68 g of NaCl (58.44 g/mol) in water and adds water to make exactly 250.0 mL of solution. What is the molarity of the solution?

  1. 18.7 M
  2. 0.0801 M
  3. 0.000320 M
  4. 0.320 MAnswer

How to solve it

  1. Moles: 4.6858.44=0.0801\frac{4.68}{58.44} = 0.0801 mol.
  2. Volume: 250.0 mL = 0.2500 L.
  3. M=0.08010.2500=0.320M = \frac{0.0801}{0.2500} = 0.320 M.

Why each choice is right or wrong

  • A. Incorrect. 18.7 M divides grams by liters without converting the mass to moles first.
  • B. Incorrect. 0.0801 M is the number of moles, 0.0801 mol, not the moles per liter of solution.
  • C. Incorrect. 0.000320 M divides by 250.0 mL instead of 0.2500 L, so the volume unit is wrong.
  • D. Correct. 4.68 g ÷ 58.44 g/mol = 0.0801 mol, and 0.0801 mol ÷ 0.2500 L = 0.320 M.

Trap Per liter of solution, not of water

Molarity uses the final volume of the solution, not the volume of water you started with. Dissolving a solute in 1.00 L of water does not make exactly 1.00 L of solution.

Explain Explain it

Task verbs: calculate, explain

For a dilution, say what stays the same (the moles of solute) and what changes (the volume, and so the concentration). Then M1V1=M2V2M_1 V_1 = M_2 V_2 follows.

For a mixture of two solutions that share an ion, count that ion's moles from both, then divide by the combined volume. A shared ion does not simply average or add its concentrations.

Worked example Medium

A student mixes 50.0 mL of 0.200 M NaCl with 50.0 mL of 0.100 M CaCl₂. Assume the volumes add and no precipitate forms.

What is the concentration of chloride ion in the mixture?

  1. 0.150 M
  2. 0.200 MAnswer
  3. 0.300 M
  4. 0.400 M

How to solve it

  1. Cl⁻ from NaCl: 0.0500×0.200=0.01000.0500 \times 0.200 = 0.0100 mol.
  2. Cl⁻ from CaCl₂: 0.0500×0.100×2=0.01000.0500 \times 0.100 \times 2 = 0.0100 mol.
  3. Total: 0.0200 mol in 0.1000 L, so [Cl−]=0.200[\text{Cl}^-] = 0.200 M.

Why each choice is right or wrong

  • A. Incorrect. 0.150 M counts one chloride per CaCl₂, but each CaCl₂ unit releases two Cl⁻ ions.
  • B. Correct. Cl⁻ from NaCl is 0.0500 L × 0.200 M = 0.0100 mol, and from CaCl₂ is 0.0500 L × 0.100 M × 2 = 0.0100 mol; 0.0200 mol ÷ 0.1000 L = 0.200 M.
  • C. Incorrect. 0.300 M adds the two concentrations, but mixing also doubles the volume, which dilutes each solution.
  • D. Incorrect. 0.400 M counts the chloride correctly but divides by only one 50.0 mL portion instead of the 100.0 mL total.

Calculate Your turn

A student must make 250.0 mL of 0.100 M copper(II) sulfate solution from a 2.00 M stock solution. Calculate the volume of stock solution needed. Then explain what the student does with that volume to make the final solution.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Calculates 12.5 mL of stock solution
  • Says the measured stock is diluted with water until the total volume of solution is 250.0 mL

Model answer

The moles of copper(II) sulfate stay the same on dilution, so M₁V₁ = M₂V₂: V₁ = (0.100 M × 250.0 mL) ÷ 2.00 M = 12.5 mL. The student measures 12.5 mL of the stock with a pipet or buret, transfers it to a 250.0 mL volumetric flask, and adds distilled water until the bottom of the meniscus reaches the 250.0 mL mark, then mixes.

Apply Apply it

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