50% NaOH Molarity Calculator (w/w to M)

A 50% (w/w) sodium hydroxide solution is approximately 18.9 M. Using the standard relationship M = (10 × %w/w × density) ÷ molar mass, with density 1.515 g/mL and molar mass 40.00 g/mol: M = (10 × 50 × 1.515) ÷ 40.00 = 18.94 M. Use the calculator below to convert any concentration or plan a dilution.

Molarity & dilution calculator

Convert any % w/w reagent to molarity, then work out how much stock to dilute. Values are editable — defaults are typical 20 °C densities; use your reagent’s certificate of analysis for exact work.

Reagent preset

Concentration (% w/w)

Density (g/mL)

Molar mass (g/mol)

18.94 M

M = (10 × 50 × 1.515) ÷ 40.00 = 18.94 M

Stock concentration (M)

Target concentration (M)

Target volume (mL)

26.4 mL of stock

V₁ = (C₂ × V₂) ÷ C₁, then make up to the target volume with solvent

Safety: when diluting concentrated acids, always add acid to water, never water to acid. Prepare in a fume hood with appropriate PPE and follow your SDS. For NaOH, dissolution is strongly exothermic — cool and add slowly.

Illustrative calculation from the standard relationship M = 10·(%w/w)·ρ ÷ M_w. Density varies with temperature and grade — verify against your reagent’s certificate of analysis and the applicable procedure.

Overview

A 50% solution of sodium hydroxide is a clear and colorless liquid. It is available is in premade format from a number of suppliers^1-3. It causes severe skin burns and eye damage. The solution may be corrosive to metals. The following are know characteristics of the solution.

Density of Sodium Hydroxide = 1.515 g/ml
Molecular weight of NaOH = 40.00 g/mol
Concentration of Sodium Hydroxide = 50% (% by mass, wt/wt)

Looking for the molar mass itself rather than this solution’s concentration? See molar mass of NaOH for the full calculation, gram-to-mole conversions, and how much to weigh for any molarity.

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Calculation

Step 1: Calculate the volume of 100 grams of Sodium Hydroxide solution.

Formula:
Density = weight / volume or
Volume = weight / density
The volume of 100 g of Sodium hydroxide solution: 100/1.515 = 66.0066 ml

The volume of 100 g of Sodium hydroxide solution is 66.0066 ml which means 50g of NaOH is present in 66.0066 ml of Sodium hydroxide solution.

Step 2: Calculate how many grams of NaOH is present in 1000 ml of Sodium hydroxide solution.

66.0066 ml of Sodium Hydroxide contains = 50 grams of NaOH
1 ml of Sodium Hydroxide will contain = 50/66.0066 grams of NaOH
1000 ml of Sodium Hydroxide will contain = 1000 x 50/66.0066 = 757.50 grams of NaOH
1000 ml of Sodium hydroxide solution will contain 757.50 grams of NaOH.

40 g of NaOH is equal to 1 mole.
1 g of NaOH will be equal to 1/40 moles.
757.50 g of NaOH will be equal to = 757.50 x 1/40 = 18.9375 moles

Therefore, 1 liter of sodium hydroxide solution contains 18.9375 moles or in other words, molarity of 50% (w/w) Sodium Hydroxide is equal to18.9375 M.

Molarity of 50% (w/w) Sodium Hydroxide (NaOH) — Value & Calculator