Molarity (M) is a measure of solution concentration defined as the number of moles of solute dissolved per liter of solution. It is expressed in units of mol/L (M). A 1 M solution of NaCl contains 1 mole of NaCl (58.44 g) dissolved in enough water to make exactly 1 liter of solution. Molarity is the most commonly used concentration unit in chemistry.

How do you calculate molarity?

Molarity = Moles of solute ÷ Volume of solution in liters (M = n/V). To find moles from mass: n = mass (g) ÷ molar mass (g/mol). Combined formula: M = mass / (molar mass × volume in L). Example: dissolve 5.85 g of NaCl (molar mass 58.44 g/mol) in 500 mL (0.5 L): moles = 5.85/58.44 = 0.1 mol; M = 0.1/0.5 = 0.2 M.

What is the dilution formula?

C₁V₁ = C₂V₂, where C₁ = initial concentration, V₁ = initial volume, C₂ = final concentration, V₂ = final volume. This formula works because the number of moles of solute stays constant during dilution. Example: dilute 100 mL of 2 M HCl to 0.5 M: V₂ = (2 × 100) / 0.5 = 400 mL total. Add 300 mL of solvent to 100 mL of stock.

What is the difference between molarity and molality?

Molarity (M) = moles of solute per liter of solution (mol/L). Molality (m) = moles of solute per kilogram of solvent (mol/kg). Molarity depends on volume (which changes with temperature), while molality depends on mass (temperature-independent). For dilute aqueous solutions, molarity ≈ molality. For concentrated solutions or temperature-sensitive work, molality is preferred.

Molar mass is the mass of one mole (6.022 × 10²³ particles) of a substance, expressed in grams per mole (g/mol). It equals the sum of atomic masses of all atoms in the chemical formula. Water (H₂O): 2(1.008) + 15.999 = 18.015 g/mol. NaCl: 22.990 + 35.453 = 58.443 g/mol. Glucose (C₆H₁₂O₆): 6(12.011) + 12(1.008) + 6(15.999) = 180.156 g/mol.

How do you calculate mass of solute needed for a given molarity?

Mass (g) = Molarity (M) × Volume (L) × Molar Mass (g/mol). Example: prepare 250 mL of 0.1 M NaCl: mass = 0.1 × 0.25 × 58.44 = 1.461 g. Weigh out 1.461 g of NaCl, dissolve in a small amount of water, then make up to exactly 250 mL total volume in a volumetric flask.

What does 1 M mean in chemistry?

1 M (one molar) means 1 mole of solute dissolved per liter of solution. For NaCl, 1 M = 58.44 g/L. For glucose, 1 M = 180.16 g/L. In biology, concentrations are often expressed in millimolar (mM = 0.001 M) or micromolar (μM = 0.000001 M). Blood glucose is normally about 5 mM (5 mmol/L).

How do you prepare a dilute solution from a concentrated stock?

Use C₁V₁ = C₂V₂. Determine C₁ (stock concentration), desired C₂ and V₂, then solve for V₁ (volume of stock to take). Example: prepare 500 mL of 0.1 M HCl from 12 M concentrated HCl: V₁ = (0.1 × 500) / 12 = 4.17 mL. Carefully add 4.17 mL of concentrated HCl to ~400 mL water, then make up to exactly 500 mL. Always add acid to water, never water to acid.

What is percent concentration vs molarity?

Percent concentration (% w/v) = grams of solute per 100 mL of solution. Molarity = moles of solute per liter. To convert % w/v to molarity: M = (% × 10 × density) / molar mass. Example: 37% HCl (density 1.19 g/mL, MW 36.46): M = (37 × 10 × 1.19) / 36.46 = 12.1 M. Percent is practical for making solutions; molarity is used for stoichiometric calculations.

What is the concentration of pure water?

The molarity of pure water is approximately 55.5 mol/L. This is calculated from water's density (1 g/mL) and molar mass (18.015 g/mol): 1000 g/L ÷ 18.015 g/mol = 55.51 mol/L. This value is important in equilibrium constants and is often used as a standard state in thermodynamic calculations.

M = n/V · Chemistry · Lab Solutions

Molarity
Calculator

Calculate molarity, moles, mass, or volume of any solution. Includes dilution calculator (C₁V₁=C₂V₂), 20+ compound library, and step-by-step solutions.

M=n/V

Formula

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C₁V₁

=C₂V₂

20+

Compounds

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Molarity Calculator

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M = n ÷ V = mass ÷ (MM × V)

Mass of solute grams

Molar mass (MM) g/mol — or use compound above

g/mol

Volume of solution

mass = M × V(L) × MM

Target Molarity mol/L

Volume of solution

Molar mass (MM)

g/mol

V = n ÷ M

Moles of solute (n)

mol

Desired Molarity

C₁V₁ = C₂V₂

Enter any 3 values — leave one blank to solve for it.

C₁ — Initial concentration

V₁ — Initial volume

C₂ — Final concentration

V₂ — Final volume

Reference Data

Molar Masses of Common Compounds

Use these values as input for your molarity calculations.

Compound	Molar Mass (g/mol)	Formula	Category	Common Use
Water	18.015	H₂O	Solvent	Universal solvent, reference
Sodium chloride	58.443	NaCl	Salt	Saline solutions, electrolyte
Sodium hydroxide	39.997	NaOH	Base	Titrations, pH adjustment
Hydrochloric acid	36.461	HCl	Acid	Acid solutions, titrations
Sulfuric acid	98.072	H₂SO₄	Acid	Electrochemistry, synthesis
Acetic acid	60.052	CH₃COOH	Acid	Buffer solutions
Glucose	180.156	C₆H₁₂O₆	Organic	Biology, cell culture
Sucrose	342.297	C₁₂H₂₂O₁₁	Organic	Density gradients
Ethanol	46.068	C₂H₅OH	Organic	Solvent, sterilization
Potassium chloride	74.551	KCl	Salt	Electrolyte solutions
Calcium chloride	110.978	CaCl₂	Salt	Cell biology, desiccant
Sodium bicarbonate	84.007	NaHCO₃	Salt	Buffer, pH 8.3 reference
Ammonia	17.031	NH₃	Base	pH adjustment
EDTA (disodium)	336.206	C₁₀H₁₄N₂Na₂O₈	Organic	Chelating agent, biology
Phosphoric acid	97.994	H₃PO₄	Acid	Buffer systems

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Always Add Acid to Water — Never Water to Acid

When preparing dilute acid solutions (especially H₂SO₄), always add the concentrated acid slowly to water while stirring. Adding water to concentrated acid causes violent, exothermic splattering. Use a volumetric flask and make up to final volume after the solution cools. Always wear eye protection and appropriate PPE when handling concentrated acids.

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Concentration Units Compared

Molarity (M): mol/L. Changes with temperature (volume changes). Most common in chemistry labs.

Molality (m): mol/kg solvent. Temperature-independent. Used for colligative properties (boiling point elevation, freezing point depression).

% w/v: g/100 mL. Simple for practical use. Used in clinical settings (e.g., 0.9% NaCl saline = 9 g/L).

ppm / ppb: mg/L (ppm) or μg/L (ppb). Used for trace concentrations in environmental science.

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Molarity in Biology

In biology and medicine, concentrations are often in mM (millimolar) or μM (micromolar). Blood glucose is ~5 mM. Physiological saline is 154 mM NaCl (0.154 M). Cell culture media contain dozens of compounds at carefully controlled molarities.

The intracellular K⁺ concentration is ~150 mM; extracellular Na⁺ is ~145 mM. These precisely maintained gradients drive nerve impulses and cell volume regulation — making accurate molarity calculations essential in biomedical research.

How to Calculate Molarity — All Four Formulas

Molarity is the most fundamental concentration unit in chemistry. The core relationship M = n/V (moles per liter) can be rearranged to solve for any of the four key variables: molarity, moles, mass, or volume.

The Four Core Formulas

Molarity: M = n ÷ V(L) or M = mass ÷ (MM × V) Moles: n = M × V(L) Mass: mass = M × V(L) × MM Volume: V(L) = n ÷ M or V = mass ÷ (M × MM) Where: M = Molarity (mol/L) n = moles of solute (mol) V = volume of solution in Liters MM = Molar mass of solute (g/mol) Example 1 — Find molarity: Dissolve 5.85 g NaCl (MM = 58.44) in 500 mL water n = 5.85 ÷ 58.44 = 0.1001 mol M = 0.1001 ÷ 0.5 = 0.200 M Example 2 — Find mass needed: Prepare 250 mL of 0.1 M NaCl mass = 0.1 × 0.25 × 58.44 = 1.461 g

The Dilution Formula: C₁V₁ = C₂V₂

C₁V₁ = C₂V₂ C₁ = initial (stock) concentration V₁ = volume of stock solution to take C₂ = final (diluted) concentration V₂ = final total volume Moles are conserved: C₁V₁ = moles taken = C₂V₂ Solvent volume to add = V₂ − V₁ Example: Make 200 mL of 0.25 M HCl from 12 M stock V₁ = (C₂ × V₂) ÷ C₁ V₁ = (0.25 × 200) ÷ 12 = 4.17 mL of stock Add 4.17 mL of 12 M HCl → make up to 200 mL total Volume of water to add ≈ 195.83 mL

Volume Unit Conversions

1 L = 1000 mL = 10 dL = 1,000,000 μL To convert to Liters for M = n/V: mL → L: divide by 1000 (500 mL = 0.5 L) dL → L: divide by 10 (2.5 dL = 0.25 L) μL → L: divide by 1,000,000

Molarity vs Other Concentration Units

Understanding when to use molarity versus other concentration expressions is important in chemistry and biology.

Converting % w/v to Molarity

% w/v (percent weight/volume) = grams of solute per 100 mL of solution = 10 g/L. To convert to molarity: M = (10 × % w/v) ÷ Molar Mass. Example: 37% HCl (MW = 36.46 g/mol): M = (37 × 10) / 36.46 = 10.15 M. Note: for concentrated solutions, density must also be factored in.

Converting ppm to Molarity

For dilute aqueous solutions: ppm ≈ mg/L. To convert: M = ppm ÷ (Molar Mass × 1000). Example: 50 ppm NaCl (MW = 58.44): M = 50 / (58.44 × 1000) = 0.000856 M = 0.856 mM. Used in environmental chemistry and trace analysis.

Why Molarity Changes with Temperature

Molarity is based on volume, which expands when heated. A 1 M solution prepared at 20°C will have slightly lower molarity at 80°C because the same amount of solute occupies more volume. For temperature-sensitive work, molality (mol/kg solvent) is preferred since mass doesn't change with temperature. For most lab work at room temperature, this difference is negligible.

Frequently Asked Questions

Molarity (M) = moles of solute per liter of solution (mol/L). A 1 M NaCl solution contains 1 mole (58.44 g) of NaCl dissolved in enough water to make exactly 1 L total volume. It is the most common concentration unit in chemistry. Not to be confused with molality (mol per kg solvent) or normality (equivalents per liter).

M = moles ÷ volume(L). First convert mass to moles: n = mass(g) ÷ molar mass(g/mol). Then M = n ÷ V(L). Example: 5.85 g NaCl in 500 mL → n = 5.85/58.44 = 0.100 mol → M = 0.100/0.5 = 0.200 M. Use Panel 1 above — enter mass, molar mass, and volume to get molarity instantly with step-by-step working.

C₁V₁ = C₂V₂. Moles of solute are conserved during dilution. To find volume of stock needed: V₁ = (C₂ × V₂) / C₁. Example: make 500 mL of 0.1 M from 6 M stock: V₁ = (0.1 × 500) / 6 = 8.33 mL. Take 8.33 mL of stock, add water to total 500 mL. Use Panel 4 to solve for any of the four variables.

Molarity (M) = mol/L solution — volume-based, changes slightly with temperature. Molality (m) = mol/kg solvent — mass-based, temperature-independent. For dilute aqueous solutions near room temp, they're approximately equal. For precise work (boiling point elevation, freezing point depression) or at non-standard temperatures, use molality.

Molar mass = mass of 1 mole of a substance in g/mol. Equal to the sum of atomic masses in the formula. H₂O = 2(1.008) + 15.999 = 18.015 g/mol. NaCl = 22.990 + 35.453 = 58.443 g/mol. Glucose (C₆H₁₂O₆) = 180.156 g/mol. Use the compound library in our calculator to select common compounds with pre-filled molar masses.

mass = M × V(L) × MM. Example: 250 mL of 0.1 M NaCl: mass = 0.1 × 0.25 × 58.44 = 1.461 g. Weigh 1.461 g NaCl, dissolve in ~200 mL water, transfer to a 250 mL volumetric flask, and make up to the 250 mL mark. Use Panel 2 to calculate the exact mass for any compound, molarity, and volume.

1 M = 1 molar = 1 mole per liter of solution. For NaCl, 1 M = 58.44 g/L. For glucose, 1 M = 180.16 g/L. In biology, much lower concentrations are common — blood glucose ≈ 5 mM (0.005 M), physiological saline = 0.154 M NaCl, and many drug concentrations are expressed in μM or nM.

Use C₁V₁ = C₂V₂. Example: 200 mL of 0.5 M HCl from 12 M stock: V₁ = (0.5 × 200) / 12 = 8.33 mL. Safety: always add acid to water (never reverse). Add 8.33 mL of concentrated HCl to ~150 mL water in a beaker, let cool, then transfer to a 200 mL volumetric flask and make up to the mark.

% w/v = grams per 100 mL. Molarity = mol/L. Convert % to M: M = (% × 10) / MM (for solutions with density ≈ 1 g/mL). Example: 5% glucose: M = (5 × 10) / 180.16 = 0.278 M. For dense concentrated acids: M = (% × 10 × density) / MM. Clinical saline (0.9% NaCl) = (0.9 × 10) / 58.44 = 0.154 M.

Pure water molarity ≈ 55.5 mol/L. Calculated as density(1000 g/L) / molar mass(18.015 g/mol) = 55.51 M. This value is used in acid-base equilibria — pure water has [H⁺] = [OH⁻] = 10⁻⁷ M at 25°C. The concentration of water is often omitted from Kw expressions because it's essentially constant in dilute solutions.

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