# How to prepare solutions and dilutions

## Calculating a mass to be weighed

### For a mass concentration

**Example**: prepare a 100 mL solution of 1 mg/L of Na^{+} with NaCl salt

^{+}

_{NaCl}= 58.44 g/mol ; M

_{Na}= 22.99 g/mol ; M

_{Cl}= 35.45 g/mol

It can be seen here that in 1 mg of NaCl, there is in fact only 0.39 mg of Na^{+}. This ratio must therefore be taken into account when calculating the mass of salt to be weighed in order to obtain a concentration of 1mg/L of Na^{+}.

#### Calculating the mass of salt to be weighed

**m**= mass of salt to be weighed (in g)

_{salt}**M**= molar mass of salt (in g/mol)

_{salt}**M**= molar mass of the element to consider (in g/mol)

_{element}**C**= final mass concentration (in g/L)

_{f}**V**= final volume (in L)

_{f}##### Mass of NaCl to be weighed

Thus in order to obtain a 100 mL solution of 1mg/L of Na^{+}, we will need to weigh 0.254 mg of salt. In reality, weighing 0.254 mg of salt is practically impossible because the most precise scales have a precision of 0.1 mg. We will need to prepare a more concentrated solution and then dilute it in order to obtain the desired concentration level of Na^{+}.

### For a molar concentration

**Example:** prepare a 100 mL solution of 1 mol/L Na^{+} with Na_{2}SO_{4} salt

Here 1 mole of Na

_{2}SO

_{4}contains 2 moles of Na

^{+}

**m**= mass of salt to be weighed (in g)

_{salt}**M**= molar mass of salt (in g/mol)

_{salt}**n**= molar ratio between the element and the salt

_{salt}/n_{element}**C**= final molar concentration (in mol/L)

_{f}**V**= final volume (in L)

_{f}Thus, the mass of salt to be weighed in order to prepare a 1 mol/L solution of Na^{+} is :