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Molarity from Percentage Composition — %w/w and %w/v Conversions

By Aniket Bhardwaj · 2 October 2026 · Calculator/Formula Guide

Reagent bottles and NCERT problems often quote concentration as a percentage rather than molarity — "9% w/v glucose" or "36% w/w HCl". Converting between these forms is a standard Class 12 solutions-chapter skill, and it appears again in IIT-JAM and CUET-PG solution-preparation questions. The key difference between the two percentage forms is whether the "100" in the denominator refers to volume or mass — and that single distinction decides whether you need the solution's density.

%w/v — mass per volume, no density needed

%w/v means grams of solute per 100 mL of solution.

Molarity (M) = (10 × %w/v) ÷ Molar mass

The factor of 10 comes from converting "grams per 100 mL" into "grams per 1000 mL (1 L)".

%w/w — mass per mass, density is required

%w/w means grams of solute per 100 g of solution. Because this tells you nothing about volume, you must be given (or look up) the solution's density d, in g/mL, to find molarity.

Molarity (M) = (10 × d × %w/w) ÷ Molar mass

Worked example 1 — %w/v to molarity

A glucose (C₆H₁₂O₆) solution is labelled 9.0% w/v. Find its molarity.

Molar mass of glucose: C = 6 × 12.011 = 72.066; H = 12 × 1.008 = 12.096; O = 6 × 15.999 = 95.994.
Total = 72.066 + 12.096 + 95.994 = 180.156 g/mol.

M = (10 × 9.0) ÷ 180.156 = 90 ÷ 180.156 = 0.4996 mol/L ≈ 0.50 M

Worked example 2 — %w/w to molarity, using density

Concentrated hydrochloric acid is commonly supplied as 36% w/w, with a density of 1.18 g/mL. Find its molarity.

Molar mass of HCl: H = 1.008, Cl = 35.45. Total = 1.008 + 35.45 = 36.458 g/mol.

M = (10 × 1.18 × 36) ÷ 36.458 = 424.8 ÷ 36.458 = 11.65 mol/L

This matches the well-known fact that concentrated lab-grade HCl is roughly 11–12 M.

Worked example 3 — working backwards, molarity to %w/v

What %w/v is a 0.1 M NaOH solution?

Molar mass of NaOH: Na = 22.990, O = 15.999, H = 1.008. Total = 22.990 + 15.999 + 1.008 = 39.997 ≈ 40.00 g/mol.

Rearranging the %w/v formula: %w/v = (M × Molar mass) ÷ 10 = (0.1 × 40.00) ÷ 10 = 0.40% w/v

This is exactly the standard "0.4 g NaOH per 100 mL" recipe used to prepare 0.1 M NaOH in a laboratory — a useful cross-check on the formula itself.

Common mistakes that cost marks

  • Using the %w/v formula for a %w/w label, or vice versa — check which one the question actually states before choosing a formula.
  • Forgetting the factor of 10 that converts "per 100" into "per 1000" (per litre).
  • Treating %w/v and %w/w as interchangeable for concentrated solutions — for dilute aqueous solutions, density is close to 1 g/mL, so the two are nearly equal, but this stops being a safe approximation once density moves away from 1 g/mL, as it does for concentrated acids.
  • Bracket errors in the molar mass calculation — always expand every subscript, including those inside brackets, before adding.

Where this is tested

Exam / courseTypical use
CBSE Class 12 Chemistry — SolutionsConverting between concentration terms (%w/v, %w/w, molarity, molality)
JEE/NEET numericalsReagent-preparation and dilution problems quoted in percentage terms
IIT-JAM / CUET-PG ChemistrySolution-preparation and stock-solution questions
GATE ChemistryConcentration-unit conversions in analytical and physical chemistry sections

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Solutions and concentration terms are a heavily-tested Class 12 chapter. ABC Chemistry's Gurugram centre and online classes across India cover this topic in depth for boards and competitive exams — details at abcchemistry.in.