Dr. Ray's Learning Center

Molarity

General Chemistry · Expressions of Concentration

Dr. Ray’s Learning Center · covers: Molarity

  • 20–30 min
  • 7 figures
  • 1 simulator
  • 3 worked examples
  • 4 practice sets

Before this lesson:

A bottle labeled 0.200 M NaCl tells you exactly how much salt is in every portion you pour: 0.200 mol in each liter, 0.0200 mol in each 100 mL. That is why chemists, nurses and water engineers describe solutions by molarity: measure a volume and you know the moles.

Molarity links three quantities: the moles of solute, the volume of solution and the concentration. Grams enter through the molar mass, as in earlier lessons (Figure 1).

  1. startMass of solutegrams (g)
  2. ×1M\times \dfrac{1}{M}molar mass
  3. Moles of solutemol
  4. ×1V\times \dfrac{1}{V}volume of solution (L)
  5. answerMolaritymol/L = M

Going right: divide. Going left (moles or grams from a molarity): multiply by V, then by M.Volumes are always in L.

Figure 1. The molarity roadmap. Grams become moles through the molar mass, and moles become a concentration by dividing by the volume of solution in L.

Learning Objectives

After this lesson, you should be able to:

  1. Calculate molarity from the amount or mass of solute and the solution volume.
  2. Rearrange the molarity equation to find moles, mass or volume.

Key Ideas

1. Molarity Is Measured in mol/L

C=nsoluteVsolutionC = \dfrac{n_{\text{solute}}}{V_{\text{solution}}}
  • CC = molarity (mol/L, written M)
  • nsoluten_{\text{solute}} = amount of solute (mol)
  • VsolutionV_{\text{solution}} = volume of the whole solution (L)

The same 0.100 mol of salt is twice as concentrated in 0.250 L as in 0.500 L (Figure 2). Molarity is a ratio, so it does not depend on how much solution you pour out: a 10 mL sample of a 0.200 M solution is still 0.200 M.

Same moles of salt in two different volumesTwo beakers each hold 0.100 mol of NaCl, drawn as 14 dots. The left one holds 0.500 L of solution, so C = 0.200 M; the right one holds 0.250 L, so the dots are more crowded and C = 0.400 M.0.100 mol NaCl in 0.500 LC = 0.200 M0.100 mol NaCl in 0.250 LC = 0.400 MSame moles, different volume
Figure 2. Same amount, half the volume, twice the concentration. Each beaker holds 0.100 mol of NaCl (14 dots each); molarity is moles divided by liters of solution.

Key idea: molarity uses moles of solute and liters of solution. Convert grams to moles and milliliters to liters first.

2. Volume of Solution, Not of Water

The volume in the formula is the volume of the finished solution. To make 500.0 mL of a solution, dissolve the solute in less water, then add water up to the 500.0 mL mark of a volumetric flask (Figure 3). Adding the solid to 500.0 mL of water gives a volume that is not exactly 500.0 mL.

Preparing a solution in a volumetric flaskThree steps. 1: weigh 5.85 g of solute on a balance. 2: dissolve it in less than 500 mL of water in a 500 mL volumetric flask. 3: add water up to the 500.0 mL mark on the neck and mix. The final volume of solution is 500.0 mL.Making 500.0 mL of solution5.85 g1. Weigh the solute2. Dissolve in less water500.0 mL mark3. Fill to the mark, mix
Figure 3. Dissolve first, then fill to the mark. The 500.0 mL in C=nVC = \frac{n}{V} is the volume of the finished solution, set by the mark on the flask neck.

Try it: make a solution

Change the mass of solute and the volume of the flask. Find the settings of Worked Example 1, then see what doubling the mass or the volume does to the molarity.

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3. Rearranging the Molarity Equation

The same relation answers three questions. Cover the quantity you want in the triangle (Figure 4) and read off the formula:

  • C=nVC = \frac{n}{V} to find a concentration,
  • n=C×Vn = C \times V to find the moles in a volume of solution,
  • V=nCV = \frac{n}{C} to find the volume that holds a given amount.
The C, n, V triangleA triangle with n on top and C times V along the bottom. Covering C leaves n over V; covering n leaves C times V; covering V leaves n over C. Units: n in mol, V in L, C in mol/L.One relation, three formulasnCV×Want C:C = nVWant n:n = C × VWant V:V = nCn in mol, V in L, C in mol/L
Figure 4. Cover the quantity you want: the triangle gives C=nVC = \frac{n}{V}, n=C×Vn = C \times V and V=nCV = \frac{n}{C}.

When a mass is given or wanted, add one step with the molar mass: n=mMn = \frac{m}{M} or m=n×Mm = n \times M.

Worked Example 1: Molarity from Mass and Volume

Problem: A technician dissolves 5.85 g of sodium chloride (NaCl\mathrm{NaCl}) in water and makes the solution up to 500.0 mL in a volumetric flask. What is the molarity of the solution?

Step 1. List the given information.

  • mNaCl=5.85gm_{\mathrm{NaCl}} = 5.85\,\text{g}
  • MNaCl=58.44g/molM_{\mathrm{NaCl}} = 58.44\,\text{g/mol}
  • V=500.0mLV = 500.0\,\text{mL}

Step 2. Convert the volume to liters.

V=500.0mL×1L1000mL=0.5000LV = 500.0\,\text{mL} \times \dfrac{1\,\text{L}}{1000\,\text{mL}} = 0.5000\,\text{L}

Step 3. Convert the mass of NaCl to moles.

nNaCl=5.85g58.44g/mol=0.1001moln_{\mathrm{NaCl}} = \dfrac{5.85\,\text{g}}{58.44\,\text{g/mol}} = 0.1001\,\text{mol}

Step 4. Divide the moles by the volume in L.

C=0.1001mol0.5000L=0.2002mol/L≈0.200M\begin{aligned}C &= \dfrac{0.1001\,\text{mol}}{0.5000\,\text{L}} \\ &= 0.2002\,\text{mol/L} \approx 0.200\,\text{M}\end{aligned}

Final Answer: 0.200 M NaCl

  1. NaCl given5.85g5.85\,\text{g}
  2. Step 3×1mol58.44g\times \dfrac{1\,\text{mol}}{58.44\,\text{g}}grams to moles
  3. NaCl0.1001mol0.1001\,\text{mol}
  4. Step 4×10.5000L\times \dfrac{1}{0.5000\,\text{L}}moles to mol/L
  5. Molarity0.2002M≈0.200M0.2002\,\text{M} \approx 0.200\,\text{M}Final answer to 3 significant figures, set by 5.85 g
Figure 5. Worked Example 1 on the roadmap (Figure 1): divide by the molar mass, then by the volume in L.

Exam tip: the answer has three significant figures because 5.85 g has three. Keep the fourth figure in the moles (0.1001 mol) and round only at the end.

Your turn: molarity from mass and volume

Same three steps as Worked Example 1: volume in L, mass to moles, moles divided by liters.

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Worked Example 2: Mass of Solute Needed

Problem: What mass of sodium hydroxide (NaOH\mathrm{NaOH}) is needed to make 250.0 mL of 0.150 M NaOH solution?

Step 1. Convert the volume to liters.

V=250.0mL×1L1000mL=0.2500LV = 250.0\,\text{mL} \times \dfrac{1\,\text{L}}{1000\,\text{mL}} = 0.2500\,\text{L}

Step 2. Find the moles with n=C×Vn = C \times V.

nNaOH=0.150mol/L×0.2500L=0.03750moln_{\mathrm{NaOH}} = 0.150\,\text{mol/L} \times 0.2500\,\text{L} = 0.03750\,\text{mol}

Step 3. Convert moles to mass. MNaOH=40.00g/molM_{\mathrm{NaOH}} = 40.00\,\text{g/mol}.

mNaOH=0.03750mol×40.00g/mol=1.500g≈1.50g\begin{aligned}m_{\mathrm{NaOH}} &= 0.03750\,\text{mol} \times 40.00\,\text{g/mol} \\ &= 1.500\,\text{g} \approx 1.50\,\text{g}\end{aligned}

Final Answer: 1.50 g NaOH

  1. Molarity given0.150M0.150\,\text{M}
  2. Step 2×0.2500L\times 0.2500\,\text{L}n = C × V
  3. NaOH0.03750mol0.03750\,\text{mol}
  4. Step 3×40.00g/mol\times 40.00\,\text{g/mol}moles to grams
  5. Mass needed1.500g≈1.50g1.500\,\text{g} \approx 1.50\,\text{g}Final answer to 3 significant figures, set by 0.150 M
Figure 6. Worked Example 2 runs the roadmap backwards: multiply by the volume, then by the molar mass.

Lab note: weigh the 1.50 g of NaOH, dissolve it in about 150 mL of water in a 250 mL volumetric flask, let it cool (dissolving NaOH releases heat), then fill to the mark and mix.

Your turn: mass of solute needed

Same method as Worked Example 2: moles from molarity and volume, then grams from the molar mass.

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Worked Example 3: Volume That Contains a Given Mass

Problem: A cell-culture recipe calls for 12.0 g of glucose (C6H12O6\mathrm{C_6\!H_{12}\!O_6}). What volume of 0.250 M glucose solution, in mL, contains that mass?

Step 1. Convert the mass to moles. MC6H12O6=180.16g/molM_{\mathrm{C_6\!H_{12}\!O_6}} = 180.16\,\text{g/mol}.

n=12.0g180.16g/mol=0.06661moln = \dfrac{12.0\,\text{g}}{180.16\,\text{g/mol}} = 0.06661\,\text{mol}

Step 2. Divide the moles by the molarity, V=nCV = \dfrac{n}{C}.

V=0.06661mol0.250mol/L=0.2664LV = \dfrac{0.06661\,\text{mol}}{0.250\,\text{mol/L}} = 0.2664\,\text{L}

Step 3. Convert liters to milliliters.

V=0.2664L×1000mL1L=266.4mL≈266mLV = 0.2664\,\text{L} \times \dfrac{1000\,\text{mL}}{1\,\text{L}} = 266.4\,\text{mL} \approx 266\,\text{mL}

Final Answer: 266 mL of glucose solution

  1. Glucose given12.0g12.0\,\text{g}
  2. Step 1×1mol180.16g\times \dfrac{1\,\text{mol}}{180.16\,\text{g}}grams to moles
  3. Glucose0.06661mol0.06661\,\text{mol}
  4. Step 2×1L0.250mol\times \dfrac{1\,\text{L}}{0.250\,\text{mol}}divide by the molarity
  5. Volume0.2664L0.2664\,\text{L}
  6. Step 3×1000mL1L\times \dfrac{1000\,\text{mL}}{1\,\text{L}}liters to mL
  7. Volume266.4mL≈266mL266.4\,\text{mL} \approx 266\,\text{mL}Final answer to 3 significant figures
Figure 7. Worked Example 3: grams to moles, then divide by the molarity to get liters of solution.

Exam tip: check the size. 0.250 M means 0.250 mol in 1000 mL; you need about a quarter of that amount (0.0666 mol), so a bit more than a quarter of a liter is right.

Your turn: volume that contains a given mass

Same method as Worked Example 3: grams to moles, divide by the molarity, then liters to milliliters.

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Common Mistakes

Common mistake: Dividing by milliliters. Molarity is in mol/L, so the volume must be in L. In Worked Example 1, 0.1001mol500.0=0.000200\frac{0.1001\,\text{mol}}{500.0} = 0.000200, a thousand times too small. Convert mL to L first.

Common mistake: Using grams instead of moles. 5.85g0.5000L=11.7g/L\frac{5.85\,\text{g}}{0.5000\,\text{L}} = 11.7\,\text{g/L} is a mass concentration in g/L, not molarity. Divide the mass by the molar mass first.

Common mistake: Using the volume of water instead of the volume of solution. Dissolve first, then fill to the mark: the volume in C=nVC = \frac{n}{V} is the final volume of solution (Key Idea 2).

Common mistake: Stopping at moles, or dividing by the molar mass. When grams are asked for, finish with m = n × M (multiply, do not divide).

Common mistake: Rounding too early. Keep 4 significant figures in intermediate values and round only the final answer to 3.

Key Takeaway

Molarity, in mol/L, is the moles of solute divided by the liters of solution. Rearrange C=nVC = \frac{n}{V} for moles or volume, use the molar mass for grams, and always work in L.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

This summary collects every practice problem you try on this page and suggests what to review next.

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Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. The simulator and the practice problems are audited; see the lesson audit log.