Dilution formula M₁V₁ = M₂V₂: Exercises — Chemistry deep dive

Every solution below spells out WHAT we do, WHY we do it, and a sanity check.

Level 1 — Recognition (can you spot the formula and plug in?)

L1·Q1

You take $100 mL$ of $3 M$ sugar solution and add water until the total volume is $300 mL$ . What is the new molarity?

Recall Solution

What we know: $M_{1} = 3$ , $V_{1} = 100$ , $V_{2} = 300$ , find $M_{2}$ . Why the formula applies: We only added water (solvent) — no sugar was added or removed, so moles are conserved and $M_{1} V_{1} = M_{2} V_{2}$ holds. $M_{2} = \frac{M _{1} V _{1}}{V _{2}} = \frac{3 \times 100}{300} = 1 M$ Sanity check: Volume tripled → concentration must fall to one-third. $3/3 = 1$ . ✔ Notice we kept both volumes in mL — they cancel, so litres are unnecessary.

L1·Q2

$25 mL$ of $0.8 M$ NaCl is diluted to $100 mL$ . Final concentration?

Recall Solution

What we know: $M_{1} = 0.8$ , $V_{1} = 25$ , $V_{2} = 100$ , find $M_{2}$ . Why the formula applies: We diluted with water only — no NaCl was added or destroyed, so moles are conserved and $M_{1} V_{1} = M_{2} V_{2}$ holds. $M_{2} = \frac{M _{1} V _{1}}{V _{2}} = \frac{0.8 \times 25}{100} = \frac{20}{100} = 0.2 M$ Sanity check: $25 \to 100$ mL is a 4× volume increase, so concentration drops 4×: $0.8/4 = 0.2$ . ✔

L1·Q3

What final volume $V_{2}$ must you dilute $40 mL$ of $5 M$ solution to, to reach $2 M$ ?

Recall Solution

What we solve for: $V_{2}$ . Why the formula applies: dilution with pure solvent conserves moles, so $M_{1} V_{1} = M_{2} V_{2}$ . $V_{2} = \frac{M _{1} V _{1}}{M _{2}} = \frac{5 \times 40}{2} = 100 mL$ Sanity check: Concentration fell by a factor of $5/2 = 2.5$ , so volume must rise by 2.5×: $40 \times 2.5 = 100$ . ✔ Note this is the total final volume, not the water added.

Level 2 — Application (find water added; solve for the stock volume)

L2·Q1

How much water must be added to $150 mL$ of $4 M$ KOH to make it $1 M$ ?

Recall Solution

Why the formula applies: you add only water (solvent) to the KOH, so the moles of KOH are conserved and $M_{1} V_{1} = M_{2} V_{2}$ holds; the formula returns the total final volume. Step 1 — total final volume. The formula gives $V_{2}$ , not water: $V_{2} = \frac{M _{1} V _{1}}{M _{2}} = \frac{4 \times 150}{1} = 600 mL$ Step 2 — water added. Subtract the starting volume: $V_{water} = V_{2} - V_{1} = 600 - 150 = 450 mL$ Sanity check: Concentration fell 4×, so volume rose 4×: $150 \to 600$ . ✔

L2·Q2

What volume of $12 M$ stock HCl is needed to prepare $2 L$ of $0.30 M$ HCl?

Recall Solution

Why the formula applies: you will take some stock HCl and add water up to the target volume — the added liquid is pure solvent, so the HCl moles you scooped out are simply spread into $2 L$ . Moles conserved ⇒ $M_{1} V_{1} = M_{2} V_{2}$ . What we solve for: the stock volume $V_{1}$ . $V_{1} = \frac{M _{2} V _{2}}{M _{1}} = \frac{0.30 \times 2}{12} = \frac{0.60}{12} = 0.05 L = 50 mL$ Why litres are fine here: both $V$ 's were in litres — as long as both sides match, the unit cancels. The answer comes out in that same unit (litres), which we convert to mL for convenience. Sanity check: Strong acid ( $12$ M) squeezed into weak ( $0.30$ M) → tiny stock volume. ✔

L2·Q3

$60 mL$ of $2.5 M$ glucose is diluted by adding $90 mL$ of water. Final molarity?

Recall Solution

Why the formula applies: only water is added, so glucose moles are conserved and $M_{1} V_{1} = M_{2} V_{2}$ holds — but first we must build the correct $V_{2}$ . Trap-aware first move: they gave us water added ( $90$ mL), not the final volume. Build $V_{2}$ : $V_{2} = V_{1} + V_{water} = 60 + 90 = 150 mL$ Now apply the formula: $M_{2} = \frac{M _{1} V _{1}}{V _{2}} = \frac{2.5 \times 60}{150} = \frac{150}{150} = 1 M$ Sanity check: $60 \to 150$ mL is a 2.5× increase; $2.5/2.5 = 1$ . ✔

Level 3 — Analysis (mixing, sequential dilutions, hidden conservation)

L3·Q1 — Mixing two solutions of the same solute

You mix $200 mL$ of $1.0 M$ NaCl with $300 mL$ of $0.5 M$ NaCl. What is the final molarity? (Assume volumes add.)

Recall Solution

Why the plain formula is NOT enough: here we add solute-containing solution, not pure water. So moles are still conserved, but they come from two sources. We conserve total moles, not a single $M_{1} V_{1}$ . Step 1 — moles from each part. Using $n = M \times V$ with $V$ in mL gives the count in mmol (recall the definition callout: $mol L^{- 1} \times mL = mmol$ ): $n_{1} = 1.0 \times 200 = 200 mmol, n_{2} = 0.5 \times 300 = 150 mmol$ Step 2 — total moles and total volume: $n_{total} = 200 + 150 = 350 mmol, V_{total} = 200 + 300 = 500 mL$ Step 3 — final molarity (mmol ÷ mL gives $mmol mL^{- 1} = mol L^{- 1} = M$ , because the "milli" cancels top and bottom): $M = \frac{n _{total}}{V _{total}} = \frac{350}{500} = 0.7 M$ Sanity check: Answer sits between $0.5$ and $1.0$ (a weighted average), closer to $0.5$ because there was more of it. ✔ ( $1.0 \times V_{1} + 0.5 \times V_{2}$ generalises $M_{1} V_{1}$ to a sum of $M V$ terms.)

The figure below draws this exact mixing: two shaded beakers carrying $200$ and $150$ mmol pour into one $500 mL$ beaker holding the summed $350$ mmol. Trace how the moles add (orange + teal → plum) while the volumes add independently — the final $0.7$ M is their ratio, not a sum of concentrations.

L3·Q2 — Two-step (serial) dilution

$10 mL$ of $6 M$ acid is diluted to $60 mL$ . Then $10 mL$ of that new solution is diluted to $100 mL$ . Final molarity?

Recall Solution

Why do it in two applications: each dilution conserves its own moles, so apply $M_{1} V_{1} = M_{2} V_{2}$ once per step. Step 1: $M_{a} = \frac{6 \times 10}{60} = 1 M$ Step 2 (feed $M_{a}$ as the new starting concentration): $M_{b} = \frac{M _{a} \times 10}{100} = \frac{1 \times 10}{100} = 0.1 M$ Sanity check via total factor: step 1 dilutes 6×, step 2 dilutes 10×, so overall 60×. $6/60 = 0.1$ . ✔

L3·Q3 — Solve for an unknown volume in a mix

You have $500 mL$ of $0.2 M$ KNO₃. What volume of $2.0 M$ KNO₃ must you add so the mixture becomes $0.5 M$ ? (Volumes add.)

Recall Solution

Let $x$ = mL of the $2.0$ M solution added. Conserve total moles (in mmol, since $M \times$ mL $=$ mmol): $have 0.2 \times 500 + add 2.0 \times x = final 0.5 \times (500 + x)$ $100 + 2 x = 250 + 0.5 x$ $1.5 x = 150 \Rightarrow x = 100 mL$ Sanity check: final volume $= 600$ mL, moles $= 100 + 200 = 300$ mmol, $M = 300/600 = 0.5$ . ✔

Level 4 — Synthesis (bridge to normality, ppm, and back-calculation)

L4·Q1 — Dilution in ppm

A stock solution is $500 ppm$ of fluoride. You take $20 mL$ and dilute to $250 mL$ . Final ppm?

Recall Solution

What ppm means here: "parts per million" is a concentration; in a dilute water solution the standard reading is $1 ppm = 1 mg L^{- 1}$ (one milligram of solute per litre of solution) — an "amount ÷ volume" measure, so it dilutes exactly like molarity. (In general ppm can instead mean a pure mass fraction, mg of solute per kg of solution; for dilute aqueous solutions the two nearly coincide because $1 L$ of water weighs $\approx 1 kg$ . See ppm and parts-per notation.) Why the same formula works: because ppm is amount-per-volume, adding pure water conserves the fluoride amount, so $C_{1} V_{1} = C_{2} V_{2}$ holds with $C$ in ppm. $C_{2} = \frac{C _{1} V _{1}}{V _{2}} = \frac{500 \times 20}{250} = \frac{10000}{250} = 40 ppm$ Sanity check: $20 \to 250$ mL is a 12.5× dilution; $500/12.5 = 40$ . ✔

L4·Q2 — Normality dilution

$30 mL$ of $6 N$ H₂SO₄ is diluted to $500 mL$ . Find the new normality, then state its molarity. (H₂SO₄ provides 2 equivalents per mole.)

Recall Solution

Why $N_{1} V_{1} = N_{2} V_{2}$ : normality is equivalents per litre — again amount÷volume — and equivalents are conserved on dilution just like moles. See Normality and N1V1 = N2V2. $N_{2} = \frac{N _{1} V _{1}}{V _{2}} = \frac{6 \times 30}{500} = \frac{180}{500} = 0.36 N$ Convert to molarity: for H₂SO₄, $N = 2 M$ , so $M = \frac{N _{2}}{2} = \frac{0.36}{2} = 0.18 M$ Sanity check: $30 \to 500$ mL is a $16. \overline{6}$ × dilution; $6/16.67 \approx 0.36$ . ✔

L4·Q3 — Back-calculate the stock

After diluting some unknown-volume stock to $250 mL$ you measure $0.12 M$ . You know the stock was $3 M$ . What starting volume of stock did you use, and how much water was added?

Recall Solution

Step 1 — stock volume ( $V_{1}$ ): $V_{1} = \frac{M _{2} V _{2}}{M _{1}} = \frac{0.12 \times 250}{3} = \frac{30}{3} = 10 mL$ Step 2 — water added: $V_{water} = V_{2} - V_{1} = 250 - 10 = 240 mL$ Sanity check: dilution factor $3/0.12 = 25$ ; volume rose 25× ( $10 \to 250$ ). ✔

Level 5 — Mastery (multi-stage, degenerate cases, reasoning)

L5·Q1 — Serial dilution to hit a target factor

You need to dilute a $1.0 M$ dye down to $1.0 \times 1 0^{- 3} M$ using only 10-fold steps (each step: take $10 mL$ , dilute to $100 mL$ ). How many steps are required, and what is the concentration after each?

Recall Solution

Why serial 10× steps: each step multiplies concentration by $\frac{10}{100} = \frac{1}{10}$ . After $k$ steps the concentration is $1.0 \times (1/10)^{k}$ . We need $(1/10)^{k} = 1 0^{- 3}$ , so $k = 3$ steps. $Step 1: 1.0 \to 0.1 M, Step 2: 0.1 \to 0.01 M, Step 3: 0.01 \to 0.001 M$ Sanity check: $1 0^{- 1} \cdot 1 0^{- 1} \cdot 1 0^{- 1} = 1 0^{- 3}$ . ✔ Three clean tenfold cuts.

The figure below stacks the three beakers left-to-right; watch the shading fade by a factor of ten at each arrow. This is why serial dilution reaches tiny concentrations that would be impractical to make in one step — each $\times \frac{1}{10}$ compounds.

L5·Q2 — The degenerate "no dilution" case

What does $M_{1} V_{1} = M_{2} V_{2}$ predict if you add zero solvent ( $V_{2} = V_{1}$ )? And what if you somehow doubled the volume by adding more of the same-concentration solution instead of water?

Recall Solution

Case A — zero solvent: set $V_{2} = V_{1}$ . Then $M_{1} V_{1} = M_{2} V_{1} \Rightarrow M_{2} = M_{1}$ . Concentration unchanged — the formula correctly says "no dilution, no change." A good boundary check that the algebra behaves. Case B — add equal-concentration solution (worked in full). Start with $V_{1}$ of concentration $M_{1}$ , and add another $V_{1}$ of the same concentration $M_{1}$ .

Moles present: the original beaker has $M_{1} V_{1}$ , the added beaker also has $M_{1} V_{1}$ , so total $= M_{1} V_{1} + M_{1} V_{1} = 2 M_{1} V_{1}$ .
Total volume: $V_{1} + V_{1} = 2 V_{1}$ .
Final concentration: $M_{final} = \frac{2 M _{1} V _{1}}{2 V _{1}} = M_{1}$ . Unchanged — correct, because you mixed identical solutions. Now watch the naive error explicitly. If instead you (wrongly) plug into $M_{1} V_{1} = M_{2} V_{2}$ using $V_{2} = 2 V_{1}$ (the new total volume), you get $M_{2}^{(naive)} = \frac{M _{1} V _{1}}{V _{2}} = \frac{M _{1} V _{1}}{2 V _{1}} = \frac{M _{1}}{2} .$ That predicts the concentration halved — but it did not. The naive step secretly assumed the extra $V_{1}$ carried zero solute (i.e. was water). Because the added liquid actually carried $M_{1} V_{1}$ moles, the true numerator is $2 M_{1} V_{1}$ , not $M_{1} V_{1}$ . Lesson: $M_{1} V_{1} = M_{2} V_{2}$ silently assumes the added volume carries zero solute. Break that assumption and you must go back to $\frac{\sum M _{i} V _{i}}{\sum V _{i}}$ .

L5·Q3 — Where the formula must NOT be used

$50 mL$ of $0.2 M$ HCl is mixed with $50 mL$ of $0.2 M$ NaOH. A student writes $M_{1} V_{1} = M_{2} V_{2}$ to get the "final HCl molarity." Why is this wrong, and what actually happens?

Recall Solution

Why it's wrong: $M_{1} V_{1} = M_{2} V_{2}$ assumes solute is only spread out, never consumed. Here HCl and NaOH react: $HCl + NaOH \to NaCl + H_{2} O$ . Moles of HCl are destroyed, so they are NOT conserved — the core assumption fails. (See Titration and neutralisation.) What actually happens: moles HCl $= 0.2 \times 50 = 10$ mmol; moles NaOH $= 0.2 \times 50 = 10$ mmol. Equal → they neutralise completely. Final HCl (and NaOH) molarity $= 0$ ; the solution is neutral NaCl salt water. Sanity check: equal moles of a strong acid and strong base annihilate each other; nothing acidic or basic left over. ✔

Active recall

Connections

Parent: the derivation and core formula
Molarity and concentration units — the $M = n / V$ every step rests on.
The Mole concept — moles are the conserved currency.
Normality and N1V1 = N2V2 — L4·Q2 uses this directly.
ppm and parts-per notation — L4·Q1 shows the same logic in ppm.
Titration and neutralisation — L5·Q3, where dilution logic must NOT be used.