Describe in detail how you would prepare a solution that contains 10 g of table salt (NaCl) and 500 mL of water to a precise Molarity. Make sure your include all the steps necessary to make this solution (i.e necessary calculations and the procedure you would have to carry out in the lab) & to describe what is happening to the molecules as you prepare the solution.

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28-10-2022
Chemistry

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Describe in detail how you would prepare a solution that contains 10 g of table salt (NaCl) and 500 mL of water to a precise Molarity. Make sure your include all the steps necessary to make this solution (i.e necessary calculations and the procedure you would have to carry out in the lab) & to describe what is happening to the molecules as you prepare the solution.

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TimofeyR353189 TimofeyR353189 · Answer 1 · 2022-10-28T13:30:27-04:00

The first thing to do is to dry the salt at a moderate temperature, about 40 to 50°C. A little more than 10 grams of NaCl is weighed and every few minutes it is weighed again until there is no significant change in weight. This is done because the salt may contain water and the precision of the procedure would be reduced, so the water should be removed.

Then we weigh the 10 grams of salt on a balance.

We measure 500 mL of water, preferably in a volumetric balloon for greater precision.

We mix both components and shake the mixture to dilute well the salt (Solute) in the water. The sodium (Na) and chlorine (Cl) atoms, initially bound together in the form of a crystal, are dissolved by the water molecules. Water is a solvent. In the end, we will have a homogeneous mixture, i.e. we can only see one phase in spite of having two substances.

The molarity of the mixture is calculated with the following equation:

[tex]Molarity=M=\frac{MolesOfSolute}{LitersOfSolution}[/tex]

So, we need to calculate the moles of NaCl. We will use the molar mass of NaCl.

Molar mass of NaCl = 58.44g/mol

[tex]\begin{gathered} \text{Moles of NaCl=}\frac{\text{Mass of NaCl}}{Moles\text{ of NaCl}} \\ \text{Moles of NaCl=}\frac{10g}{58.44\frac{g}{mol}} \\ \text{Moles of NaCl=}0.17\text{mol} \end{gathered}[/tex]

Volumen of solution= 500mL = 0.5L

So, the molarity will be:

[tex]\begin{gathered} M=\frac{\text{Moles of NaCl}}{LitersOfSolution} \\ M=\frac{0.17\text{mol}}{0.5L} \\ M=0.34M \end{gathered}[/tex]

The molarity of this mix will be 0.34M, and we will have a homogeneous mixture.