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# 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.