# Solutions, molarity and dilution

## Definitions and examples of how to calculate molarity and prepare dilutions

### Solutions

A solution is in chemistry defined as:

1. A homogeneous mixture composed of two or more substances.
2. The process by which a gas, liquid, or solid is dispersed homogeneously in a gas, liquid, or solid without chemical change.
3. As an example; such a substance, as dissolved sugar or salt in solution.

The substance (most often a liquid) in which another substance dissolves is called the solvent, while the dissolved substance is called the solute.

The solution assumes the characteristics of the solvent (i.e., its phase) when the solvent is the larger fraction of the mixture, as is commonly the case. From this, the solute is usually the component of a solution present in the lesser amount.

The solubility of a substance in another is not unlimited, and  how much solute you can solve in a solvent varies a lot.

The concentration of a solute in a solution is can be given as

1. the mass of solute expressed as a percentage of the mass of the whole solution,  which is the mass%
2. the volume of solute expressed as a percentage of the volume of the whole solution, which is the volume%
3. the number of moles of the solute in a given volume of the whole solution, whis is the molarity.
4. the number of moles of the solute in a given mass of the solvent, which is the molality.
5. the ratio between the number of moles of the solute  and the total number of moles of the solution, which is the molar fraction.

### Molarity

Molarity expresses the molar concentration of a compound in a solution, how many moles of the solute in a given volume of the solution:

M = n / V          [1]

n: number of moles of solute [mol]

M: Molarity of the solution [mol/L] or [M]

V: volume of the solution [L]

The unit for molarity is molar, with the symbol M: 1 M = 1 mol/L, where L refers to the volume of the whole solution. A solution with a concentration of 1 mol/L is equivalent to 1 molar (1 M).

From the definition, we can calculate the number of moles of the solute, n,:

n = M * V        [2]

### Dilution

Dilution is the prosess where a solution is added more of the solvent to decrease the concentration of the solute.

In dilution, the amount of solute does not change, the number of moles are the same before and after dilution.

If subscript "i" represents initial and "f" represents the final values of the quantities involved, we have:

ni = Mi * Vi    and   nf = Mf * Vf

ni = nf   and thus,

Mi * Vi = Mf * V               [3]

From this we can calculate the final molarity after dilution:

Mf = Mi * Vi / Vf          [4]

or we can calculate the needed final volume to reach a wanted final molarity:

Vf  = Mi * Vi / Mf          [5]

### Example 1

You add 200 gram of salt (NaCl) into enough water to make exactly 5 L of a salt water solution.  What is the molarity of the sloution?

Molweight of Na is 22,99 g/mol and molweight of Cl is 35,45 g/mol.

Molar weight of NaCl:   22.99 [g/mol] + 35.45 [g/mol] = 58.44 [g/mol]

Number of moles of NaCl:  200 [g] /58.44 [g/mol] = 3.42 [mol NaCl]

Molarity of solution: 3,42 mol NaCl / 5 L solution = 0.684 mol/L = 0.684 M

### Example 2

You want 2 liter of a 1M solution of acetic acid (CH3COOH) in water. How much 100% acetic acid do you need to add (in mol, in gram and in liter)?

1M solution means 1 mol acetic acid per liter of solution.

Moles needed to 2 liter solution:  2 [L] * 1 [mol/L] = 2 mol acetic acid

Molweight of C: 12.01 g/mol, molweight of H: 1.01 g/mol, molweight of O: 16.00 g/mol.

Molweight of acetic acid: 2*12.01 [g/mol]+ 4*1.01 [g/mol] + 2*16.00 [g/mol] = 60.06 [g/mol]

Gram needed to 2 liter solution: 2 [mol] * 60.06 [g/mol] = 120.12 g acetic acid

Density of 100% acetic acid at 20°C is 1.048 g/cm3

Volume needed to 2 liter solution: 120.12[g] / 1.048 [g/cm3] = 114.62 cm3 = 114.62 ml = 0.115 L acetic acid

Practical tip: Never start with adding water to acid! Use a container with a precise definition of 2 L. Add 1 liter water and then 0.115 L acetic acid (measured at 20°C) and rotate the container to blend the liquids.  Add carefully more water untill you reach 2 liter total volume.

### Example 3

You have 500 mL left of the solution with 1M acetic acid. Now you need 100 mL of a 0.3M solution. How can you make that without using more of the 100% acetic acid?

The simplest way is to dilute the 1M solution, which we do by adding more water to it.

From equation [3] above:

Vf = 100 mL=0.100 L   and  Mf = 0.3M = 0.3 mol/L

Mi= 1M = 1 mol/L,   while   Vi = unknown  (we dont know how much we need of the starting solution)

Vi = Vf * Mf /Mi   =  0.100 [L] * 0.3 [mol/L] / 1[mol/L] = 0.03 [L] = 30 mL of the 1M solution.

So, start with 30 ml of the 1 M solution and add water until you have exactly 100 ml of the diluted solution.

Check:

Number of moles acetic acid wanted in final solution: 0.100[L] * 0.3[mol/L] = 0.03 mol acetic acid

Number of moles from the 1 M solution: 0.03[L] * 1[mol/L] = 0.03 mol acetic acid

ni = n   (we have useed the correct amount of 1M solution

## Related Topics

• Basics - The SI-system, unit converters, physical constants, drawing scales and more
• Material Properties - Material properties for gases, fluids and solids - densities, specific heats, viscosities and more

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