EXPERIMENTAL PROBLEM n. 09 GROUP C

You have a given amount of sodium chloride (3,0 grams) in a small test tube. Your task is to put, in a second empty test tube, as many ions of potassium and chloride as they are in the 3,0 grams of NaCl, and then to compare the actual volumes of the two amounts of solids. You have also alcohol, cylinder and pipette.

Ricevete 3,0 g di cloruro di sodio in una piccola provetta. L'obiettivo del gruppo è di introdurre in un'altra provetta tanti ioni potassio e cloruro quanti ve ne sono nella provetta con l'NaCl, quindi comparare i volumi effettivi delle due quantità di solidi. Avete a disposizione anche alcool, cilindro e pipetta graduati.

1. Useful knowledge (reference data, concepts and principles)

Sodium Chloride (NaCl) Formula Weight = 3,0g

Potassium Chloride (KCl) Formula Weight = 3,8g

quantity of Na⁺ or Cl^- ions in 3,0 grams of NaCl = ...

mass of KCl with the same number of ions K⁺and Cl^- = ...

Alcohol don't dissolve ...

2. Main idea

synthetic description of the whole strategy to accomplish the task

Take 2 tubes, one is filled with sodium chloride, and the other of potash potassium chloride. Then you take 2 cylinders and are filled up to 5ml of alcohol. He throws on a sodium chloride and potassium chloride and are located on the volumes.

3. Planning

Put in a test tube 3.0 g of sodium chloride;
Place in a second test tube 3.8 g of potassium chloride;
Fill a graduated cylinder up to 5 mL of alcohol;
Add the sodium chloride and find the new volume;
Fill a graduated cylinder up to another 5 mL of alcohol;
Add the potassium chloride and find the new volume;

4. Report

- Sodium chloride = 3.0 g;

- Potassium chloride = 3.8 g;

- Container blank = 1.97 g;

- Container with potassium chloride = 5.77 g;

- Graduated cylinder with alcohol = 5 mL;

- Graduated cylinder with alcohol and sodium chloride = 6.69 6.4 mL;

- Graduated cylinder with alcohol = 5 mL;

- Graduated cylinder with alcohol and potassium chloride = 6.8 g mL;

5. Elaborazione e Interpretazione

Data elaboration-calculation from raw data

deductions and interpretations from calculated data

conclusions: answering problem request. Other outcomes.

- solid NaCl Volume = 6,4 - 5,0 = 1.4 mL sodium chloride;

- Solid KCl Volume = 6.8 - 5.0 = 1.8 mL potassium chloride.

The volume of potassium chloride with the same number of ions is slightly (about 30%) larger than the volume of sodium chloride.

Questions & Answers

Question 1

Why is the amount of KCl more than three grams?

Because the atomic massof a particle of potassium is greater than that a sodium.

Question 2

What data you need to calculate volumes from masses, without measuring?

We need to find the density for times (×) volume.

Question 3

Why are the volumes of the two salts so different?

Question 4

Why is a potassium ion bigger than a sodium ion?

Answer 4

Question 5

Which is the largest between:

a) a bromide ion and a chloride ion

b) a cesium ion and a cesium atom

c) a sodium ion and a magnesium ion

d) a iodine atom and a iodide ion

e) an oxygen atom and a sulphur atom

Question 6