• These are used to designate the amount of solute dissolved in a given quantity of solvent or quantity of solution.

• In a solution, we have solute and solvent.

• The solute is the one found in the smallest amount in the solution.

• The solvent is the one found in the biggest amount in the solution.

• Chemistry requires the concentration of solutions expressed quantitatively:

Molarity (M)

M = moles of solute / volume of solution in liters

Molality (m)

m = moles of solute / kg of solvent

Molar Fraction

XX = moles of solute / total moles of solution

M/M % (mass/mass %)

mass of solute / total mass of solution *100

V/V % (volume/volume %)

volume of solute / total volume of solution *100

ppm (parts per million)

mass of solute (mg) / total mass of solution (Kg)

ppb (parts per billion)

mass of solute (μg) / total mass of solution (Kg)

Example:

A solution is prepared by dissolving 4.35 g of glucose (C6H12O6) in 25,0 mL of water. Calculate the molality of glucose in the solution. Water has a density of 1,00 g/mL

m = moles of solute/ kg of solvent

Glucose’s molar mass → 180,15 g/mol

we need to transform 4,35 g into moles

4,35 g / 180,15g/mol = 0,02414 moles of glucose

we need to transform 25 mL into L to obtain Kg because L=Kg

25mL/1000mL = 0,025 L → 0,025 Kg

now we substitute our values in the formula

0,02414 moles of glucose / 0,025 Kg = 0,9656m