# Concentrations of aqueous solutions

## Reminders

### Volumic mass of the solution ($\rho_S $)

The volumic mass of the solution is the ratio of the mass of the solution by its volume:
$\rho_S$ $ =$ $\frac{m_S}{V_S}$

For example,
$\rho_{S}$ $ =$ $ 0.95\frac{g}{cm^{3}}$
means that $1 cm^3$ of the solution S has a mass of 0.9 g
Indeed, in this case, we have:
$\rho_{S}$ =
$\frac{0.95}{1}$ =
$0.95\frac{g}{cm^{3}} $

### Number of moles of solute ($n_{so}$)

The number of moles of solute is the ratio of the mass of the solute by its molar mass:
$ n_{so}$ $ =$ $\frac{m_{S}}{M_{S}} $

For example $18 \;g$ of glucose $( C_6H_{12}O_6 )$ are
$ n_{glucose}$ =
$\frac{18}{6\cdot 12 + 12\cdot 1 + 6\cdot 16}$ =
$0.1\; mol$

## Definitions

### Percentage of solute ($ \%_{so}$)

The percentage of the solute is the ratio of the mass of the solute by the mass of the solution, multiplied by $ 100$:
$\%_{so}$ $ =$ $\frac{m_{so}\cdot100}{M_S} $

For example, $ \%_{so}$ $ = $ $24$ means that $100\; g$ of this solution contained $24\; g$ of solute $so$
Indeed, in this case, we have:
$ \%_{so}$ =
$\frac{24\cdot 100}{100}$ =
24

### Molar concentration (Molarity) of the solute ($ c_{so} $)

The molarity of the solute is the ratio of the number of moles of solute per volume of solution.
$ c_{so}$ $ =$ $\frac{n_{so}}{V_S} $

Remember that we agreed to call $ c_{so} $ the molarity of the supposed undissociated solute (before dissociation), the remaining molarity after dissociation being called $ [so] $:
For example:
$ c_{so}$ =
$2\frac{mol}{L}$ =
$2\frac{mol}{L}$ or $2\;M$
(not to be confused with the molar mass M!)
It means that $1\; L$ of the solution $S$ contains $2\; mol$ of solute $so$
Indeed, in this case, we have:
$ c_{so}$ =
$\frac{2}{1}$ =
$2\frac{mol}{L}$

## Calculations

Each colored triangle corresponds to one of the above formulas:
In each triangle, a quantity may be calculated if we know the other two.
The diagram allows to switch from one triangle to another.