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Making up solutions

15 min

  • Calculate molarity or reagent mass using the solute molar mass
  • Calculate normality for the stated reaction
  • Calculate a percent solution on its stated mass/volume, volume/volume, or mass/mass basis
  • Calculate the stock volume required for a target solution

Read the full reference

Try first

Try first

How many grams of NaOH (molar mass 40.00 g/mol) make 500 mL of a 0.100 mol/L solution?

The next section explains it.

Right. The next section explains why.

The next section explains it.

The next section explains it.

Get the idea

Reagent recipes state a concentration in one of three ways, and each has its own arithmetic. The work is mostly keeping units straight.

Molarity

Molarity is moles of solute per liter of finished solution. The mass to weigh out is:

mass (g) = mol/L × volume (L) × molar mass (g/mol)

The volume goes in as liters, so 250 mL is 0.250 L. Missing that step makes the answer a thousand times too large.1,2

Normality

Normality is equivalents per liter, and it belongs to a reaction:

normality = molarity × n

Here n counts the protons, hydroxide ions, electrons or charges that take part in the stated reaction. The formula alone does not set n.1

  • Sulfuric acid gives up two protons when fully neutralized, so 1 mol/L is 2 N.
  • Phosphoric acid has three protons. When only one reacts, n is 1 and the normality equals the molarity.

Percent solutions

Percent solutions come in three forms, and the label says which:3

FormMeans5% example
w/vgrams of solute per 100 mL of solution5 g in each 100 mL
v/vmilliliters of solute per 100 mL of solution5 mL in each 100 mL
w/wgrams of solute per 100 g of solution5 g in each 100 g

A w/w percent needs the solution's density before it can say anything about a volume. Concentrated acids are labeled this way, so their strength per milliliter is density × mass fraction. In these problems the specific gravity stands in for the density in g/mL.3

Stock volume

A working solution is often made by diluting a concentrated stock. With both concentrations in the same unit:

C₁V₁ = C₂V₂, so V₁ = C₂V₂ ÷ C₁

  • V₁ is the stock to measure.
  • V₂ is the final volume of the whole solution.2 The stock is brought up to V₂ in total. Adding a full V₂ of water to the stock would overshoot the volume and leave the solution weaker than planned.
  • With concentrated acid, the acid goes slowly into water, with ventilation and protective equipment.4
References
  1. Bishop ML, Fody EP, Van Siclen C, Mistler JM, Moy M. Clinical Chemistry: Principles, Techniques, and Correlations. 9th ed. Jones & Bartlett Learning; 2023.
  2. Flowers P, Theopold K, Langley R, Robinson WR. Molarity. In: Chemistry 2e. OpenStax; 2019. Accessed September 23, 2026. https://openstax.org/books/chemistry-2e/pages/3-3-molarity
  3. Flowers P, Theopold K, Langley R, Robinson WR. Other units for solution concentrations. In: Chemistry 2e. OpenStax; 2019. Accessed September 23, 2026. https://openstax.org/books/chemistry-2e/pages/3-4-other-units-for-solution-concentrations
  4. National Institute for Occupational Safety and Health. Engineering Controls Database: acids and alkalis. Accessed September 23, 2026.

Watch one

You need 250.0 mL of 0.500 mol/L nitric acid.

  • The stock is 70.0% w/w HNO₃ with a specific gravity of 1.42.
  • The molar mass of HNO₃ is 63.01 g/mol.

How much stock do you measure, and how do you make up the solution?

  1. Find the HNO₃ per milliliter of stock: 1.42 g/mL × 0.700 = 0.994 g/mL, or 994 g/L.

    A w/w percent is grams per 100 g, so density turns it into grams per volume.

  2. Convert to mol/L: 994 g/L ÷ 63.01 g/mol = 15.775 mol/L, kept unrounded.

    C₁V₁ = C₂V₂ needs both concentrations in the same unit.

  3. Solve for V₁: (0.500 mol/L × 250.0 mL) ÷ 15.775 mol/L = 7.924 mL.

    The stock volume carries the moles the working solution needs.

  4. Round to 7.92 mL.

    Round once, at the end, to the precision of the data.

  5. Add the 7.92 mL of acid slowly to about 200 mL of water in a 250 mL volumetric flask, then bring it to the 250.0 mL mark with water.

    V₂ is the final volume of the whole solution, stock included.

Measure 7.92 mL of stock and bring it to 250.0 mL in total.

Your turn

Problem 1 of 3

What mass of NaCl makes 250 mL of 0.200 mol/L solution? Use molar mass 58.44 g/mol and report three significant figures.

Incorrect. 0.200 × 0.250 = 0.0500 mol. Multiply those moles by 58.44 g/mol to obtain grams.

Correct. 0.200 mol/L × 0.250 L × 58.44 g/mol = 2.922 g, reported as 2.92 g.

Incorrect. This is ten times the required mass. Convert 250 mL to 0.250 L before multiplying.

Hint
  1. Molarity is moles per liter, so change 250 mL to liters first.
  2. Find the moles: 0.200 mol/L × 0.250 L.
  3. Then turn moles into grams with the molar mass.

Review Molarity

Problem 2 of 3

A 0.200 mol/L phosphoric acid (H₃PO₄) solution is used in a reaction in which only one of its three protons reacts. What is its normality for that reaction?

Hint
  1. Normality = molarity × n.
  2. n counts only the protons that react in this reaction.
Show the answer

0.2 eq/L

One proton reacts, so n = 1 and the normality is 0.200 mol/L × 1 = 0.200 eq/L (0.200 N). Using all three protons from the formula would give 0.600 N, three times too high for this reaction.

Review Normality

Problem 3 of 3

Concentrated sulfuric acid is labeled 96.0% w/w H₂SO₄ with a specific gravity of 1.84. How many grams of H₂SO₄ are in 100.0 mL of it?

Show the answer

177 g

The 100.0 mL weighs 100.0 mL × 1.84 g/mL = 184 g, and 96.0% of that is H₂SO₄: 184 g × 0.960 = 176.6 g, about 177 g.

Review Percent solutions

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