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Molar Concentration & Solution Chemistry Calculator
Compute molar concentrations, plan dilutions, and explore gas solubility with unit-aware tools built for laboratory precision.
Ensure molar mass is in g/mol and volume is the final solution volume.
Amount of Substance (n)
Represents how many particles are present in a sample. It is measured in moles (mol) and directly links microscopic particles to macroscopic laboratory quantities.
Applications: stoichiometric calculations, expressing solution concentration, and connecting gas volume to chemical equations.
Avogadro's constant (NA)
One mole of any substance contains approximately 6.022 × 1023 elementary entities (atoms, molecules, ions, or electrons). Units: mol⁻¹.
Molar volume (Vm)
The volume occupied by one mole of substance. For ideal gases at standard temperature and pressure (273.15 K and 101.325 kPa), Vm equals 22.4 L/mol.
Significance: equal moles of different ideal gases occupy identical volumes under the same conditions (Avogadro's law).
Molar concentration (c)
Defines the amount of solute per unit volume of solution in mol/L. It is the laboratory standard for expressing solution strength: c = n / V.
Mass m ↔ Amount n: n = m / M (M in g/mol) Amount n ↔ Gas volume V: V = n × Vₘ (Vₘ = 22.4 L/mol at STP) Amount n ↔ Particles N: N = n × Nₐ (Nₐ = 6.022 × 10²³ mol⁻¹)
Tip: sketch a star diagram with amount of substance (n) in the center and arrows toward mass, volume, and particles to keep the relationships straight.
Always check unit consistency—convert mass (g ↔ kg) and volume (mL ↔ L) before substituting values.
Dilution (C₁·V₁ = C₂·V₂)
The amount of solute stays constant before and after dilution. Start with concentrated stock solutions to conserve reagents.
For mass concentration: ρ₁·V₁·w₁ = ρ₂·V₂·w₂, where ρ is density and w is solute mass fraction.
Solution Mixing
When combining solutions of the same solute without reaction, apply C = (C₁·V₁ + C₂·V₂) / (V₁ + V₂).
Mixing different solutes? Track each component separately and account for any volume change (e.g., concentrated sulfuric acid with water).
Gas Dissolution Workflow
- Amount of gas: n = V₀ / 22.4 mol.
- Total solution volume: V = (M × n + 1000) / ρ L.
- Molar concentration: c = n / V = 1000ρV₀ / (M V₀ + 22400) mol/L.
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