# Solutions, molarity and dilution

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

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### Solutions

A **solution** is in chemistry defined as:

- A homogeneous mixture composed of two or more substances.
- The process by which a gas, liquid, or solid is dispersed homogeneously in a gas, liquid, or solid without chemical change.
- 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

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

See also Solubility product constants and Mixtures, Solutions and Suspensions. ** **

### 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:

n_{i} = M_{i} * V_{i} and n_{f} = M_{f} * V_{f}

n_{i} = n_{f} and thus,

M_{i} * V_{i} = M_{f} * V_{f } [3]

From this we can calculate the final **molarity after dilution**:

M_{f} = M_{i} * V_{i} / V_{f} [4]

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

V_{f} = M_{i} * V_{i} / M_{f} [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 (CH_{3}COOH) 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/cm^{3}

**Volume** needed to 2 liter solution: 120.12[g] / 1.048 [g/cm^{3}] = 114.62 cm^{3} = 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:

V_{f} = 100 mL=0.100 L and M_{f} = 0.3M = 0.3 mol/L

M_{i}= 1M = 1 mol/L, while V_{i} = unknown (we dont know how much we need of the starting solution)

V_{i} = V_{f} * M_{f} /M_{i }= 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*

n_{i} = n_{f } (we have useed the correct amount of 1M solution) _{ }

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