# 3.1 Solubility

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## Experiments and solubility

Two grade 10 learners, Siphiwe and Ann, wish to separately investigate the solubility of potassium chloride at room temperature. They follow the list of instructions shown below, using the apparatus that has been given to them:

Method:

1. Determine the mass of an empty, dry evaporating basin using an electronic balance and record the mass.
2. Pour 50 ml water into a 250 ml beaker.
3. Add potassium chloride crystals to the water in the beaker in small portions.
4. Stir the solution until the salt dissolves.
5. Repeat the addition of potassium chloride (steps a and b) until no more salt dissolves and some salt remains undissolved.
6. Record the temperature of the potassium chloride solution.
7. Filter the solution into the evaporating basin.
8. Determine the mass of the evaporating basin containing the solution that has passed through the filter (the filtrate) on the electronic balance and record the mass.
9. Ignite the Bunsen burner.
10. Carefully heat the filtrate in the evaporating basin until the salt is dry.
11. Place the evaporating basin in the desiccator (a large glass container in which there is a dehydrating agent like calcium sulphate that absorbs water) until it reaches room temperature.
12. Determine the mass of the evaporating basin containing the dry cool salt on the electronic balance and record the mass.

On completion of the experiment, their results were as follows:

 Siphiwe's results Ann's results Temperature ( ${}^{0}$ C) 15 26 Mass of evaporating basin (g) 65.32 67.55 Mass of evaporating basin + salt solution (g) 125.32 137.55 Mass of evaporating basin + salt (g) 81.32 85.75
1. Calculate the solubility of potassium chloride, using the data recorded by
1. Siphiwe
2. Ann A reference book lists the solubility of potassium chloride as 35.0 g per 100 ml of water at 25 ${}^{0}$ C.
3. Give a reason why you think Ann and Siphiwe each obtained results different from each other and the value in the reference book.
2. Siphiwe and Ann now expand their investigation and work together. They now investigate the solubility of potassium chloride at different temperatures and in addition they examine the solubility of copper (II) sulfate at these same temperatures. They collect and write up their results as follows: In each experiment we used 50 ml of water in the beaker. We found the following masses of substance dissolved in the 50 ml of water. At 0 ${}^{0}$ C, mass of potassium chloride is 14.0 g and copper sulphate is 14.3 g. At 10 ${}^{0}$ C, 15.6 g and 17.4 g respectively. At 20 ${}^{0}$ C, 17.3 g and 20.7 g respectively. At 40 ${}^{0}$ C, potassium chloride mass is 20.2 g and copper sulphate is 28.5 g, at 60 ${}^{0}$ C, 23.1 g and 40.0 g and lastly at 80 ${}^{0}$ C, the masses were 26.4 g and 55.0 g respectively.
1. From the record of data provided above, draw up a neat table to record Siphiwe and Ann's results.
2. Identify the dependent and independent variables in their investigation.
3. Choose an appropriate scale and plot a graph of these results.
4. From the graph, determine:
1. the temperature at which the solubility of copper sulphate is 50 g per 50 ml of water.
2. the maximum number of grams of potassium chloride which will dissolve in 100 ml of water at 70 ${}^{0}$ C.
(IEB Exemplar Paper 2, 2006)

## Summary

• In chemistry, a solution is a homogenous mixture of a solute in a solvent.
• A solute is a substance that dissolves in a solvent. A solute can be a solid, liquid or gas.
• A solvent is a substance in which a solute dissolves. A solvent can also be a solid, liquid or gas.
• Examples of solutions include salt solutions, metal alloys, the air we breathe and gases such as oxygen and carbon dioxide dissolved in water.
• Not all solutes will dissolve in all solvents. A general rule is: like dissolves like . Solutes and solvents that have similar intermolecular forces are more likely to dissolve.
• Polar and ionic solutes will be more likely to dissolve in polar solvents, while non-polar solutes will be more likely to dissolve in polar solvents.
• Solubility is the extent to which a solute is able to dissolve in a solvent under certain conditions.
• Factors that affect solubility are the quantity of solute and solvent , temperature , the intermolecular forces in the solute and solvent and other substances that may be in the solvent.

## Summary exercise

1. Give one word or term for each of the following descriptions:
1. A type of mixture where the solute has completely dissolved in the solvent.
2. A measure of how much solute is dissolved in a solution.
3. Forces between the molecules in a substance.
2. For each of the following questions, choose the one correct answer from the list provided.
1. Which one of the following will readily dissolve in water?
1. I ${}_{2}$ (s)
2. NaI(s)
3. CCl ${}_{4}$ (l)
4. BaSO ${}_{4}$ (s)
(IEB Paper 2, 2005)
2. In which of the following pairs of substances will the dissolving process happen most readily?
 Solute Solvent A S ${}_{8}$ H ${}_{2}$ O B KCl CCl ${}_{4}$ C KNO ${}_{3}$ H ${}_{2}$ O D NH ${}_{4}$ Cl CCl ${}_{4}$
(IEB Paper 2, 2004)
3. Which one of the following three substances is the most soluble in pure water at room temperature? Hydrogen sulfide, ammonia and hydrogen fluoride
4. Briefly explain in terms of intermolecular forces why solid iodine does not dissolve in pure water, yet it dissolves in xylene (a non-polar, organic liquid) at room temperature. (IEB Paper 2, 2002)

#### Questions & Answers

during a snooker competition ,a 200g ball A m moving with velocity va collide head on with a identical ball B that was at rest.A after the collision ball A remains at rest wile ball B moves on with a velocity of 4m/s? With what speed was ball a moving before the collision
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how to calculate normal force
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how to calculate wavelength
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Hello. How does a real gas behave under low temperature and high pressure?
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does a vector quantity include force and distance?
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yes
Itumeleng
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Keaobaka
vector is the physical quantity with magnitude and direction Scalar is the Physical quantity with magnitude only
Bobo
Newton's second law of motion
Thelma Reply
Newton second law motin
Shife
Newton's second law of motion: When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the force and inversely proportional to the mass of the object.
Bobo
newtons third law of motion
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down
Fabzeey
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The ratio of the sine of the angle of incidence in one medium to the sine of the angle of refraction in the other medium is constant
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who was the first person to discover nuclears bomb
Ismael Reply
how to calculate resistance in grade 11
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last year memo in 2018 June exam
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Mike
yes
Ismael
hello
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Patricia
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Patricia
states the Newton's second law of motion
Shallin Reply
In words it says:" when a net force is applied to an object of mass, It accelerates in the direction of the net force. The acceleration is directly proportional to the net force and inversely proportional to the mass".
Buhle
And in symbols: Fnet = ma
Buhle
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hlw
manu
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Mzondi
90°
Boss
yes obviously 90 degree
manu
fnet=ma
Fabzeey
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Madzivha Reply
Hydrogen bonds are the strongest intermolecular forces and London forces are the weakest. Hydrogen bonds exist between polar molecules ( they are a special case of dipole-dipole forces ), while London forces occur between non-polar molecules.
Kagiso
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Andrew Reply
It's when you have equivalent forces going different directions then your resultant will be equal to zero
Temosho
describe what john's experiment proves about water molecules?
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