## Chemistry Tutorial

*Determine the mass of MgCl2 needed to prepare 1 L solution at 0,5 M Cl.*

#### Intro

To prepare a certain amount (1 L) of solution (MgCl2) with a specific concentration (0,5 M or mol/Litre) of chloride, we will convert units from moles/L to mass, just like we would convert moles to mass.

#### Sample Problem

Determine the mass of magnesium chloride (MgCl2) needed to prepare a 1 L solution of MgCl2 at 0,5 mol Cl/L of solution.

#### Solution

First, note the molar mass (g/mol) of chloride (Cl) and magnesium (Mg) which can be found in the periodic table of your manuel (at the begginning or in the appendix at the end).

**Cl: 35,45 g/mol**

**Mg : 24,305 g/mol**

MgCl2 : 24,305 g/mol + (2 x 35,45 g/mol) = 24,305 g/mol + 70,90 g/mol = 95,205 g/mol

To calculate, start with the information you have : the concentration you want.

= 0,5 mol Cl/L solution x 1 mol MgCl2 / 2 mol Cl x 95,205 g MgCl2 / 1 mol MgCl2

Remember that the answer need the right units. Simplify the units to find what is left:

**= 23,80 g MgCl2 per 1 L of solution**

Therefore, you will need to mesure about 23,80 g of magnesium chloride MgCl2.

*To prepare the solution, first add about 900 mL of distilled water to your 1 L glass balloon. Then slowly pour your mesured 23,80 g of MgCl2 while stirring. Once all the MgCl2 is added, fill the balloon with distilled water while stirring to disolve and mix the solution evenly.*

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