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Dilution and mixing questions look like arithmetic, yet they catch students who reach for an average or forget that the final volume has changed. One idea solves all of them: count the moles of each species, then divide by the final volume. This post applies that idea to five kinds of question, with checked numbers, and ends with a routine.
What do NCERT and the syllabus say?
The NCERT Class 11 chapter Some Basic Concepts of Chemistry (Reprint 2026-27) defines molarity as moles of solute per litre of solution. It makes 0.2 M NaOH by taking 200 mL of 1 M NaOH and diluting to 1 litre, then gives the general formula M1 × V1 = M2 × V2. It adds the reason: the moles of NaOH "remained the same" because only the amount of water changed.
The JEE (Main) 2026 syllabus lists molarity under Solutions and stoichiometry under basic concepts, and the NEET (UG) 2026 syllabus does the same. The CBSE Chemistry curriculum for 2026-27 says students will prepare standard solutions by weighing for the titration practicals. Check your own exam year's documents.
Case 1: How much water do you add to dilute?
You have 250 mL of 2.0 M HCl and need 0.50 M.
Moles of HCl = 2.0 × 0.250 = 0.500 mol.
Final volume = 0.500/0.50 = 1.0 L.
Water to add = 1000 − 250 = 750 mL.
Read the question carefully. "Diluted to 1 L" means the final volume is 1 L. "Diluted with 1 L of water" means roughly 1.25 L.
Case 2: What happens when you mix two solutions of the same solute?
Mix 200 mL of 0.50 M NaCl with 300 mL of 0.20 M NaCl.
Moles: 0.50 × 0.200 = 0.100 mol, and 0.20 × 0.300 = 0.060 mol. Total 0.160 mol.
Volume: 0.500 L.
Final molarity = 0.160/0.500 = 0.32 M.
Averaging the two molarities gives 0.35 M, which is wrong because the volumes differ. The answer is a volume-weighted mean, M = (M1V1 + M2V2)/(V1 + V2), and it must lie between the two starting values.
Case 3: How do you hit a target molarity?
You need 500 mL of 0.40 M solution from stocks of 1.0 M and 0.25 M. Let x mL come from the 1.0 M stock.
Moles: 1.0x + 0.25(500 − x) = 0.40 × 500 = 200 (in millimoles, with volumes in mL).
0.75x = 75, so x = 100 mL.
Take 100 mL of 1.0 M and 400 mL of 0.25 M.
Check: (100 + 100)/500 = 0.40 M.
Case 4: Which ions are in a mixed salt solution?
Mix 100 mL of 0.10 M CaCl₂ with 100 mL of 0.20 M NaCl. Count each ion, not each salt.
Ca²⁺: 0.010 mol → 0.010/0.200 = 0.050 M.
Na⁺: 0.020 mol → 0.10 M.
Cl⁻: 2 × 0.010 from CaCl₂ plus 0.020 from NaCl = 0.040 mol → 0.20 M.
Forgetting the 2 in CaCl₂ gives 0.15 M for chloride, a common slip.
Case 5: What if the two solutions react?
Mix 50 mL of 0.20 M HCl with 30 mL of 0.25 M NaOH.
H⁺: 0.010 mol. OH⁻: 0.0075 mol.
They neutralise mole for mole, leaving 0.0025 mol H⁺ in 80 mL.
[H⁺] = 0.0025/0.080 = 0.031 M, so pH = −log(0.03125) ≈ 1.51.
With a diprotic acid, count the protons. 50 mL of 0.10 M H₂SO₄ also supplies 0.010 mol H⁺, so the same NaOH leaves the same excess, if you assume both protons react.
Do volumes always add?
Not exactly. The NCERT Class 12 chapter Solutions says that for an ideal solution ΔmixV = 0, so the volume equals the sum of the volumes of the components. Real non-ideal mixtures can shrink or expand on mixing. For dilute aqueous solutions the error is small, which is why exam questions assume additive volumes unless they give a density.
Molarity also changes with temperature, because volume does. For conversions between molarity, molality and mole fraction, see our guide to concentration terms and conversions.
Which mistakes cost marks?
Averaging molarities instead of adding moles.
Mixing mL and L in the same line. Pick one and stay with it; millimoles with mL works well.
Confusing "diluted to" with "diluted with".
Counting salts instead of ions when asked for an ion's concentration.
Forgetting the reaction when an acid meets a base, or the second proton of H₂SO₄.
A practice routine
Day 1: three dilution questions, writing moles before and after on separate lines.
Day 2: three two-solution mixes, each with the "between the two" check.
Day 3: two target-molarity problems, solved by the moles equation and checked by substitution.
Day 4: two ion-counting and two acid-base mixes, including one with H₂SO₄.
If the same step keeps going wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Start now: redo Case 2 with 100 mL of each solution and predict the answer before you calculate it.
