Solution Chemistry Calculator: Molarity, Dilution and pH

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how to calculate molarity

How to calculate the molarity (mol/L) of a solution: formula and examples

Molarity answers a single question: how many moles of solute fit in each liter of solution. The tool works it out from mass and volume, or converts from % w/w, ppm and g/L, and shows each step with your numbers.

From mass to mol/L, and from the label to mol/L

The definition is c = n ÷ V, with n in moles and V in liters of finished solution. Since n = m ÷ M, the bench formula is c = m ÷ (M × V). For sodium chloride (M = 58.44 g/mol), 5.844 g in 1 L gives 0.1000 mol, that is, 0.1000 mol/L. The most common slip is using the water volume instead of the final volume: the solution is measured up to the mark of the volumetric flask, after the solid has dissolved.
Concentrated reagents come with a content in % w/w and a density, not a molarity. The bridge is c = % × 10 × d ÷ M, with d in g/mL. Hydrochloric acid at 37 % with d = 1.19 g/mL and M = 36.458 g/mol holds 37 × 10 × 1.19 ÷ 36.458 = 12.08 mol/L. Sulfuric acid at 98 % with d = 1.84 g/mL holds 98 × 10 × 1.84 ÷ 98.072 = 18.39 mol/L. Without the density this conversion cannot be done, because % w/w divides by solution mass and mol/L divides by volume.
Other units follow the same path. In dilute water, 1 mg/L is practically 1 ppm, and 50 mg/L of NaCl is 50 ÷ 1000 ÷ 58.44 = 8.56 × 10⁻⁴ mol/L. Normality is molarity times the number of equivalents: 0.5 mol/L of H2SO4 is 1 N in full neutralization. Finally, molality (moles per kg of solvent) hardly differs from molarity in dilute solutions but departs from it a lot in concentrated ones: 0.1 mol/L NaCl (d = 1.00 g/mL) is 0.1006 mol/kg, while 98 % H2SO4 is 18.39 mol/L and about 500 mol/kg.

Molarity worked step by step

NaCl: 5.844 g in 1 L

Input: NaCl, m = 5.844 g, V = 1 L
Expected output: 0.1000 mol/L

n = 5.844 ÷ 58.44 = 0.1000 mol (CIAAW 2024, M = 58.4398).

NaOH: 4.00 g in 100 mL

Input: NaOH, m = 4.00 g, V = 0.100 L
Expected output: 1.000 mol/L

M = 39.997 g/mol; n = 0.1000 mol; 0.1000 ÷ 0.100 L.

Glucose: 18.0 g in 500 mL

Input: C6H12O6, m = 18.0 g, V = 0.500 L
Expected output: 0.200 mol/L

M = 180.156 g/mol; n = 0.0999 mol; 0.0999 ÷ 0.500 L = 0.1998.

Commercial HCl 37 %

Input: HCl, 37 % m/m, d = 1.19 g/mL
Expected output: 12.08 mol/L

37 × 10 × 1.19 ÷ 36.458.

Commercial H2SO4 98 %

Input: H2SO4, 98 % m/m, d = 1.84 g/mL
Expected output: 18.39 mol/L

98 × 10 × 1.84 ÷ 98.072; in normality, 36.77 N.

Commercial HNO3 65 %

Input: HNO3, 65 % m/m, d = 1.39 g/mL
Expected output: 14.34 mol/L

65 × 10 × 1.39 ÷ 63.012.

Full tool FAQ

How do I calculate the molarity of a solution?

Molarity is moles of solute per liter of solution: c = m ÷ (M × V). For 5.844 g of NaCl (58.44 g/mol) in 1 L of solution: 5.844 ÷ 58.44 = 0.1000 mol, so 0.1000 mol/L. V is the final solution volume, measured to the mark of the flask, not the volume of water used.

What is the difference between molarity and molality?

How do I convert % w/w to mol/L?

How much should I weigh to prepare a solution, with hydrate and purity?

How do I use C1V1 = C2V2 and what does a 1:10 dilution mean?

How do I calculate the pH of a weak acid and when does the 5 % approximation hold?

Why is the pH of 1e-8 mol/L HCl not 8?

How does a buffer work and when does Henderson-Hasselbalch fail?

How do I read a titration curve and choose the indicator?

What are the limits of the calculations and what should the tool not be used for?

Frequently asked questions

Are molarity and mass concentration the same thing?

No. Mass concentration is solute mass per liter (g/L); molarity is moles per liter. They are linked by the molar mass: g/L = mol/L × M. NaCl at 0.1000 mol/L equals 5.844 g/L.

Does molarity change with temperature?

Yes, slightly, because the solution volume expands while the moles stay the same. Molality, which uses the solvent mass, does not change. For most bench work the difference is smaller than the glassware uncertainty.

Can I use the rounded exam masses?

Yes: the tool has a rounded atomic mass option (H = 1, C = 12, O = 16, Na = 23, Cl = 35.5). With it NaCl is 58.5 g/mol and 5.85 g in 1 L gives exactly 0.1 mol/L.

Next steps

The same solution in every unit
Unit	Value
Molarity (mol/L)	0.100 mol/L
Millimolarity (mmol/L)	100 mmol/L
Normality (eq/L)	0.100 eq/L
Molality (mol/kg)	0.101 mol/kg
Mole fraction (x)	0.00181
Mass concentration (g/L)	5.84 g/L
Milligrams per liter (mg/L)	5840 mg/L
% mass/volume (g/100 mL)	0.584 % m/v
% volume/volume	needs solute density
% mass/mass	0.584 % m/m
Mass fraction (titer)	0.00584
Parts per million (ppm, mg/kg)	5840 ppm
Parts per billion (ppb, µg/kg)	5.84 × 10⁶ ppb

The same solution in every unit

Unit

Value

Molarity (mol/L)

0.100 mol/L

Millimolarity (mmol/L)

100 mmol/L

Normality (eq/L)

0.100 eq/L

Molality (mol/kg)

0.101 mol/kg

Mole fraction (x)

0.00181

Mass concentration (g/L)

5.84 g/L

Milligrams per liter (mg/L)

5840 mg/L

% mass/volume (g/100 mL)

0.584 % m/v

% volume/volume

needs solute density

% mass/mass

0.584 % m/m

Mass fraction (titer)

0.00584

Parts per million (ppm, mg/kg)

5840 ppm

Parts per billion (ppb, µg/kg)

5.84 × 10⁶ ppb

From mass to mol/L, and from the label to mol/L

The definition is c = n ÷ V, with n in moles and V in liters of finished solution. Since n = m ÷ M, the bench formula is c = m ÷ (M × V). For sodium chloride (M = 58.44 g/mol), 5.844 g in 1 L gives 0.1000 mol, that is, 0.1000 mol/L. The most common slip is using the water volume instead of the final volume: the solution is measured up to the mark of the volumetric flask, after the solid has dissolved.

Concentrated reagents come with a content in % w/w and a density, not a molarity. The bridge is c = % × 10 × d ÷ M, with d in g/mL. Hydrochloric acid at 37 % with d = 1.19 g/mL and M = 36.458 g/mol holds 37 × 10 × 1.19 ÷ 36.458 = 12.08 mol/L. Sulfuric acid at 98 % with d = 1.84 g/mL holds 98 × 10 × 1.84 ÷ 98.072 = 18.39 mol/L. Without the density this conversion cannot be done, because % w/w divides by solution mass and mol/L divides by volume.

Other units follow the same path. In dilute water, 1 mg/L is practically 1 ppm, and 50 mg/L of NaCl is 50 ÷ 1000 ÷ 58.44 = 8.56 × 10⁻⁴ mol/L. Normality is molarity times the number of equivalents: 0.5 mol/L of H2SO4 is 1 N in full neutralization. Finally, molality (moles per kg of solvent) hardly differs from molarity in dilute solutions but departs from it a lot in concentrated ones: 0.1 mol/L NaCl (d = 1.00 g/mL) is 0.1006 mol/kg, while 98 % H2SO4 is 18.39 mol/L and about 500 mol/kg.

Molarity worked step by step

NaCl: 5.844 g in 1 L

Input: NaCl, m = 5.844 g, V = 1 L
Expected output: 0.1000 mol/L

n = 5.844 ÷ 58.44 = 0.1000 mol (CIAAW 2024, M = 58.4398).

NaOH: 4.00 g in 100 mL

Input: NaOH, m = 4.00 g, V = 0.100 L
Expected output: 1.000 mol/L

M = 39.997 g/mol; n = 0.1000 mol; 0.1000 ÷ 0.100 L.

Glucose: 18.0 g in 500 mL

Input: C6H12O6, m = 18.0 g, V = 0.500 L
Expected output: 0.200 mol/L

M = 180.156 g/mol; n = 0.0999 mol; 0.0999 ÷ 0.500 L = 0.1998.

Commercial HCl 37 %

Input: HCl, 37 % m/m, d = 1.19 g/mL
Expected output: 12.08 mol/L

37 × 10 × 1.19 ÷ 36.458.

Commercial H2SO4 98 %

Input: H2SO4, 98 % m/m, d = 1.84 g/mL
Expected output: 18.39 mol/L

98 × 10 × 1.84 ÷ 98.072; in normality, 36.77 N.

Commercial HNO3 65 %

Input: HNO3, 65 % m/m, d = 1.39 g/mL
Expected output: 14.34 mol/L

65 × 10 × 1.39 ÷ 63.012.

Full tool FAQ

How do I calculate the molarity of a solution?

What is the difference between molarity and molality?

How do I convert % w/w to mol/L?

How much should I weigh to prepare a solution, with hydrate and purity?

How do I use C1V1 = C2V2 and what does a 1:10 dilution mean?

How do I calculate the pH of a weak acid and when does the 5 % approximation hold?

Why is the pH of 1e-8 mol/L HCl not 8?

How does a buffer work and when does Henderson-Hasselbalch fail?

How do I read a titration curve and choose the indicator?

What are the limits of the calculations and what should the tool not be used for?

Frequently asked questions

Are molarity and mass concentration the same thing?

No. Mass concentration is solute mass per liter (g/L); molarity is moles per liter. They are linked by the molar mass: g/L = mol/L × M. NaCl at 0.1000 mol/L equals 5.844 g/L.

Does molarity change with temperature?

Can I use the rounded exam masses?

Yes: the tool has a rounded atomic mass option (H = 1, C = 12, O = 16, Na = 23, Cl = 35.5). With it NaCl is 58.5 g/mol and 5.85 g in 1 L gives exactly 0.1 mol/L.

The same solution in every unit
Unit	Value
Molarity (mol/L)	0.100 mol/L
Millimolarity (mmol/L)	100 mmol/L
Normality (eq/L)	0.100 eq/L
Molality (mol/kg)	0.101 mol/kg
Mole fraction (x)	0.00181
Mass concentration (g/L)	5.84 g/L
Milligrams per liter (mg/L)	5840 mg/L
% mass/volume (g/100 mL)	0.584 % m/v
% volume/volume	needs solute density
% mass/mass	0.584 % m/m
Mass fraction (titer)	0.00584
Parts per million (ppm, mg/kg)	5840 ppm
Parts per billion (ppb, µg/kg)	5.84 × 10⁶ ppb

The same solution in every unit

Unit

Value

Molarity (mol/L)

0.100 mol/L

Millimolarity (mmol/L)

100 mmol/L

Normality (eq/L)

0.100 eq/L

Molality (mol/kg)

0.101 mol/kg

Mole fraction (x)

0.00181

Mass concentration (g/L)

5.84 g/L

Milligrams per liter (mg/L)

5840 mg/L

% mass/volume (g/100 mL)

0.584 % m/v

% volume/volume

needs solute density

% mass/mass

0.584 % m/m

Mass fraction (titer)

0.00584

Parts per million (ppm, mg/kg)

5840 ppm

Parts per billion (ppb, µg/kg)

5.84 × 10⁶ ppb