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}} $
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}$ = $\frac{18}{6\cdot 12 + 12\cdot 1 + 6\cdot 16}$ = $0.1\; mol$
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
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}$ = $\frac{2}{1}$ = $2\frac{mol}{L}$
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.