Solutions

Class 10 Unit 9 Science

UNIT-9

SOLUTIONS

Learning Objectives

After studying this lesson, students will be able to:

• Define solution.
• Recognize the types of solutions.
• Analyse the factors influencing solubility.
• Explain the various modes of expression of concentration of solution.
• Calculate the solubility of solutes in solvents.
• Correlate the hydrated salts and anhydrous salts.
• Distinguish between deliquescent and hygroscopic substances.

INTRODUCTION

You have learnt about mixtures in your lower classes. Most of the substances that we encounter in our daily life are mixtures of two or more substances. The substances present in a mixture may exist in one or more physical state. For example, when we burn wood, the smoke released is a mixture of solid carbon and gases like CO₂, CO, etc. In some cases of mixtures, their components can be separated easily whereas in some other cases they cannot be.

Consider the two mixtures, one which contains salt and water, and the another which contains sand and water. Water is the one of the components in both the mixtures. In the first case salt dissolves in water. In the second case the sand does not dissolve in water. Sand in water can be separated by filtration but salt cannot be separated as it dissolves in water to form a homogeneous mixture. This kind of homogenous mixtures are termed as "solutions".

9.1 SOLUTIONS IN DAY-TO-DAY LIFE

One of the naturally existing solutions is sea water. We cannot imagine life on earth without sea water. It is a mixture of many dissolved salts. Another one is air. It is a mixture of gases like nitrogen, oxygen, carbon dioxide and other gases. All the life forms on the earth are associated with solutions. Plants take solutions of nutrients for their growth from the soil. Most of the liquids found in human body including blood, lymph and urine are solutions. Day to day human activities like washing, cooking, cleaning and few other activities involve the formation of solutions with water. Similarly, the drinks what we take, like fruit juice, aerated drinks, tea, coffee etc. are also solutions. Therefore, the ability of water to form solutions is responsible for sustenance of life.

9.2 COMPONENTS OF SOLUTIONS

We know that, a solution is a homogeneous mixture of two or more substances. In a solution, the component which is present in lesser amount (by weight), is called solute and the component, which is present in a larger amount (by weight) is called solvent. The solute gets distributed uniformly throughout the solvent and thus forming the mixture homogeneous. So, the solvent acts as a dissolving medium in a solution. The process of uniform distribution of solute into solvent is called dissolution.

A solution must at least be consisting of two components (a solute and a solvent). Such solutions which are made of one solute and one solvent (two components) are called binary solutions. e.g. On adding copper sulphate crystals to water, it dissolves in water forming a solution of copper sulphate. It contains two components i.e. one solute- copper sulphate and one solvent-water So it is a binary solution. Similarly, a solution may contain more than two components. For example if salt and sugar are added to water, both dissolve in water forming a solution. Here two solutes are dissolved in one solvent. Such kind of solutions which contain three components are called ternary solutions.

9.3 Types of Solutions

9.3.1 Based on the physical state of the solute and the solvent

We know that substances normally exist in three physical states (phases) i.e., solid, liquid and gas. In binary solutions, both the solvent and solute may exist in any of these physical states. But the solvent constitutes the major part of the solution. Its physical state is the primary factor which determine the characteristics of the solution. Therefore, there are different types of binary solutions as listed in Table 9.1.

9.3.2 Based on the type of solvent

Most of the substances are soluble in water. That is why, water is called as 'Universal solvent". However some substances do not dissolve in water. Therefore, other solvents such as ethers, benzene, alcohols etc., are used to prepare a solution. On the basis of type of solvent, solutions are classified into two types. They are aqueous solutions and non-aqueous solutions.

a) Aqueous solution

The solution in which water acts as a solvent is called aqueous solution. Example: Common salt in water, Sugar in water, Copper sulphate in water etcetera.

b) Non–Aqueous solution

The solution in which any liquid, other than water, acts as a solvent is called nonaqueous solution. Solvent other than water is referred to as non-aqueous solvent. Generally, alcohols, benzene, ethers, carbon disulphide, acetone, etcetera are used as non-aqueous solvents. Examples for non-aqueous solutions: Sulphur dissolved in carbon disulphide; Iodine dissolved in carbon tetrachloride.

9.3.3 Based on the amount of solute

The amount of the solute that can be dissolved in the given amount of solvent is limited under any given conditions. Based on the amount of solute, in the given amount of solvent, solutions are classified into the following types:

(i) Saturated solution

A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution. Example: 36 gram of sodium chloride in 100 gram of water at 25°C forms saturated solution. Further addition of sodium chloride, leave it undissolved.

(ii) Unsaturated solution

Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature. Example: 10 gram or 20 gram or 30 gram of Sodium chloride in 100 gram of water at 25°C forms an unsaturated solution.

(iii) Super saturated solution

Supersaturated solution is one that contains more solute than the saturated solution at a given temperature. Example: 40 gram of sodium chloride in 100 gram of water at 25°C forms super saturated solution. This state can be achieved by altering any other conditions like temperature, pressure. Super saturated solutions are unstable, and the solute is reappearing as crystals when the solution is disturbed.

9.3.4 Concentrated and dilute solutions

It is another kind of classification of unsaturated solutions. It expresses the relative concentration of two solutions with respect to their solutes present in the given amount of the solvent. For example, you are given two cups of tea. When you taste them, you feel that one is sweeter than the other. What do you infer from it? The tea which sweet more contains higher amount of sugar than the other. How can you express your observation? You can say that the tea is stronger. But a chemist would say that it is 'concentrated'. When we compare two solutions having same solute and solvent the one which contains higher amount of solute per the given amount of solvent is said to be 'concentrated solution' and the another is said to be 'dilute solution'.

Differentiating solutions as dilute and concentrated is a qualitative representation. It does not imply the quantity of the solute. This difference is observed by means of some physical characteristics such as colour, density, etcetera.

Activity 1

Look at the following pictures. Label them as dilute and concentrated solution and justify your answer.

9.4 Solubility

Usually, there is a limit to the amount of solute that can be dissolved in a given amount of solvent at a given temperature. When this limit is reached, we have a saturated solution and any excess solute that is added, simply resides at the bottom of the solution. The extent of dissolution of a solute in a solvent can be better explained by its solubility.

Solubility is a measure of how much of a solute can be dissolved in a specified amount of a solvent. Solubility is defined as the number of grams of a solute that can be dissolved in 100 gram of a solvent to form its saturated solution at a given temperature and pressure. For example, 36 gram of sodium chloride need to be dissolved in 100 gram of water to form its saturated solution at 25°C. Thus the solubility of sodium chloride in water is 36 gram at 25°C.

Formula: Solubility = (mass of the solute × 100) / mass of the solvent

Table 9.2 shows solubilities of some common substances in water at 25°C:

Name of the solute	Formula of the solute	Solubility (gram per 100 gram water)
Calcium carbonate	CaCO₃ (solid)	0.0013
Sodium chloride	NaCl(solid)	36
Ammonia	NH₃ (gas)	48
Sodium hydroxide	NaOH(solid)	80
Glucose	C₆H₁₂O₆ (solid)	91
Sodium bromide	NaBr (solid)	95
Sodium iodide	NaI (solid)	184

9.4.1 Factors affecting solubility

There are three main factors which govern the solubility of a solute. They are:

(i) Nature of the solute and solvent

The nature of the solute and solvent plays an important role in solubility. Although water dissolves an enormous variety of substances, both ionic and covalent, it does not dissolve everything. The phrase that scientists often use when predicting solubility is "like dissolves like." This expression means that dissolving occurs when similarities exist between the solvent and the solute. For example: Common salt is a polar compound and dissolves readily in polar solvent like water. Non-polar compounds are soluble in non-polar solvents. For example, Fat dissolved in ether. But non-polar compounds, do not dissolve in polar solvents; polar compounds do not dissolve in non-polar solvents.

(ii) Effect of Temperature

Solubility of Solids in Liquid: Generally, solubility of a solid solute in a liquid solvent increase with increase in temperature. For example, a greater amount of sugar will dissolve in warm water than in cold water. In endothermic process, solubility increases with increase in temperature. In exothermic process, solubility decreases with increase in temperature.

Solubility of Gases in liquid: Do you know why is it bubbling when water is boiled? Solubility of gases in liquid decrease with increase in temperature. Generally, water contains dissolved oxygen. When water is boiled, the solubility of oxygen in water decreases, so oxygen escapes in the form of bubbles. Aquatic animals live more in cold regions because, more amount of dissolved oxygen is present in the water of cold regions. This shows that the solubility of oxygen in water is more at low temperatures.

(iii) Effect of Pressure

Effect of pressure is observed only in the case of solubility of a gas in a liquid. When the pressure is increased, the solubility of a gas in liquid increases. The common examples for solubility of gases in liquids are carbonated beverages, That is soft drinks, household cleaners containing aqueous solution of ammonia, formalin-aqueous solution of formaldehyde, etcetera.

More to Know

The effect of pressure on the solubility of a gas in liquid is given by Henry's law. It states that, the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution at a definite temperature.

9.5 Concentration of a Solution

So far, we discussed what is a solution? what does it consist of and its types. Most of the chemical reactions take place in solutions form So it is essential to quantify the solute in solvent to study the reactions. To quantify the solute in a solution, we can use the term "concentration". Concentration of a solution may be defined as the amount of solute present in a given amount of solution or solvent. Quantitatively, concentration of a solution may be expressed in different methods. But here, we shall discuss percentage by mass ( percentage of mass) and percentage by volume (percentage of volume).

9.5.1 Mass percentage

Mass percentage of a solution is defined as the percentage by mass of the solute present in the solution. It is mostly used when solute is solid and solvent is liquid.

Formula: Mass percentage = (mass of the solute × 100) / mass of the solution

Mass percentage = (mass of the solute × 100) / (mass of the solute + mass of the solvent)

For example: 5% sugar solution (by mass) means 5 gram of sugar in 95 gram of water. Hence it is made of 100gram of solution. Usually, mass percentage is expressed as w/w (weight / weight); mass percentage is independent of temperature.

9.5.2 Volume percentage

Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of the solution.

Formula: Volume percentage = (Volume of the solute × 100) / Volume of the solution

Volume percentage = (Volume of the solute × 100) / (Volume of the solute + Volume of the solvent)

For example, 10% by volume of the solution of ethanol in water, means 10 ml of ethanol in100 ml of solution (or 90 ml of water). Usually volume percentage is expressed as v/v (volume by volume). It is used when both the solute and solvent are liquids. Volume percentage decreases with increases in temperature, because of expansion of liquid.

You can notice that in the commercial products that we come across in our dailylife such as a solution of syrups, mouth wash,antiseptic solution, household disinfectants etc.,the concentration of the ingredients is expressed as v/v. Similarly, in ointments, antacid, soaps,etc., the concentration of solutions are expressed as w/w.

9.6 Hydrated salts and Water of Crystallization

When ionic substances are dissolved in water to make their saturated aqueous solution, their ions attract water molecules which then attached chemically in certain ratio. This process is called hydration. These ionic substances crystallize out from their saturated aqueous solution with a definite number of molecules of water. The number of water molecules found in the crystalline substance is called water of crystallization. Such salts are called hydrated salts. On heating these hydrated crystalline salts, they lose their water of crystallization and become amorphous or lose their colour(if they are coloured).

Table 9.3 shows some common hydrated salts:

Common Name	IUPAC Name	Molecular Formula
Blue Vitriol	Copper(II) sulphate pentahydrate	CuSO₄.5H₂O
Epsom Salt	Magnesium sulphate heptahydrate	MgSO₄.7H₂O
Gypsum	Calcium sulphate dihydrate	CaSO₄.2H₂O
Green Vitriol	Iron (II) sulphate heptahydrate	FeSO₄.7H₂O
White Vitriol	Zinc sulphate heptahydrate	ZnSO₄.7H₂O

9.6.1 Copper sulphate pentahydrate CuSO₄.5H₂O (Blue vitriol)

The number of water molecules in blue vitriol is five. So its water of crystallization is 5. When blue coloured copper sulphate crystals are gently heated, it loses its five water molecules and becomes colourless anhydrous copper sulphate.

If you add few drops of water or allow it to cool, the colourless anhydrous salt again turns back into blue coloured hydrated salt.

9.6.2 Magnesium sulphate heptahydrate MgSO₄.7H₂O (Epsom salt)

Its water of crystallization is 7. When magnesium sulphate heptahydrate crystals are gently heated, it loses seven water molecules,and becomes anhydrous magnesium sulphate.

Chemical Equation: MgSO₄.7H₂O ⇌ MgSO₄ + 7H₂O

(Magnesium sulphate heptahydrate) → (Anhydrous Magnesium sulphate)

While heating rightward reaction; while cooling leftward reaction

If you add few drops of water or allow it to cool, the colourless anhydrous salt again turns back into hydrated salt.

9.7 Hygroscopy

Certain substances, when exposed to the atmospheric air at ordinary temperature,absorb moisture without changing their physical state. Such substances are called hygroscopic substances and this property is called hygroscopy. Hygroscopic substances are used as drying agents.

Examples:

1. Conc.Sulphuric acid (H₂SO₄)

2. Phosphorus Pentoxide (P₂O₅)

3. Quick lime (CaO)

4. Silica gel (SiO₂)

9.8 Deliquescence

Certain substances which are so hygroscopic, when exposed to the atmospheric air at ordinary temperatures, absorb enough water and get completely dissolved. Such substances are called deliquescent substances and this property is called deliquescence. Deliquescent substances lose their crystalline shape and ultimately dissolve in the absorbed water forming a saturated solution.

Deliquescence is maximum when:

1) The temperature is low

2) The atmosphere is humid

Examples: Caustic soda (NaOH), Caustic potash(KOH) and Ferric chloride (FeCl₃)

Table 9.3 Difference between hygroscopic substances and deliquescence:

Hygroscopic substances	Deliquescence substances
When exposed to the atmosphere at ordinary temperature, they absorb moisture and do not dissolve.	When exposed to the atmospheric air at ordinary temperature, they absorb moisture and dissolve.
Hygroscopic substances do not change its physical state on exposure to air.	Deliquescent substances change its physical state on exposure to air.
Hygroscopic substances may be amorphous solids or liquids.	Deliquescent substances are crystalline solids.

9.9 Problems Based on Solubility and Percentage by Mass and Volume

1. Problems based on solubility

1) 1.5 gram of solute is dissolved in 15 gram of water to form a saturated solution at 298 K. Find out the solubility of the solute at the temperature.

Mass of the solute = 1.5 gram

Mass of the solvent = 15 gram

Solubility = (Mass of the solute × 100) / Mass of the solvent

Solubility of the solute = (1.5 × 100) / 15 = 10 gram

2) Find the mass of potassium chloride would be needed to form a saturated solution in 60 gram of water at 303 K? Given that solubility of the KCl is 37 per 100 gram at this temperature.

Mass of potassium chloride in 100 gram of water in saturated solution = 37 gram

Mass of potassium chloride in 60 gram of water in saturated solution = (37 × 60) / 100 = 22.2 gram

3) What is the mass of sodium chloride that would be needed to form a saturated solution in 50gram of water at 30°C. Solubility of sodium chloride is 36 gram at 30°C?

At 30°C, 36gram of sodium chloride is dissolved in 100gram of water.

Therefore Mass of sodium chloride that would be needed for 100 gram of water = 36 gram

Mass of sodium chloride dissolved in 50g of water = (36 × 50) / 100 = 18 gram

4) The Solubility of sodium nitrate at 50°C and 30°C is 114 gram and 96 gram respectively. Find the amount of salt that will be thrown out when a saturated solution of sodium nitrate containing 50 gram of water is cooled from 50°C to 30°C?

Amount of sodium nitrate dissolved in 100 gram of water at 50°C is 114 gram

Amount of sodium nitrate dissolving in 50g of water at 50°C = (114 × 50) / 100 = 57 gram

Similarly amount of sodium nitrate dissolving in 50gram of water at 30°C = (96 × 50) / 100 = 48 gram

Amount of sodium nitrate thrown when 50g of water is cooled from 50°C to 30°C = 57 - 48 = 9 gram

II. Problem based on mass percentage

1) A solution was prepared by dissolving 25 gram of sugar in 100 gram of water. Calculate the mass percentage of solute.

Mass of the solute = 25 gram

Mass of the solvent = 100 gram

Mass percentage = (mass of the solute × 100) / mass of the solution

Mass percentage = (mass of the solute × 100) / (mass of the solute + mass of the solvent)

= (25 × 100) / (25 + 100) = (25 × 100) / 125 = 20%

2) 16 grams of NaOH is dissolved in 100grams of water at 25°C to form a saturated solution. Find the mass percentage of solute and solvent.

Mass of the solute (NaOH) = 16 gram

Mass of the solvent H₂O = 100 gram

(i) Mass percentage of the solute

Mass percentage = (mass of the solute × 100) / (mass of the solute + mass of the solvent)

= (16 × 100) / (16 + 100) = 1600 / 116 = 13.79

Mass percentage of the solute = 13.79%

Mass percentage of solvent = 100 - (Mass percentage of the solute) = 100 - 13.79 = 86.21%

3) Find the amount of urea which is to be dissolved in water to get 500 gram of 10% w/w aqueous solution?

Mass percentage w/w = (mass of the solute × 100) / mass of the solution

10 = (mass of the urea × 100) / 500

Mass of urea = 50 gram

III. Problem based on Volume - volume percentage

1. A solution is made from 35 ml of Methanol and 65 ml of water. Calculate the volume percentage.

Volume of the ethanol = 35 ml

Volume of the water = 65ml

Volume percentage = (Volume of the solute × 100) / Volume of the solution

Volume percentage = (Volume of the solute × 100) / (Volume of the solute + Volume of the solvent)

= (35 × 100) / (35 + 65) = (35 × 100) / 100 = 35%

2. Calculate the volume of ethanol in 200 ml solution of 20% v/v aqueous solution of ethanol.

Volume of aqueous solution = 200ml

Volume percentage = (Volume of solute × 100) / Volume of solution

Volume percentage = 20%

20 = (Volume of ethanol × 100) / 200

Volume of ethanol = (20 × 200) / 100 = 40 ml

Points to Remember

• A solution is a homogeneous mixture of two or more substances.
• An aqueous solution is a solution in which the solvent is water.
• A non-aqueous solution is a solution in which the solvent is a liquid, other than water
• A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
• An unsaturated solution is one that contains less solute than the saturated solution at a given temperature.
• A supersaturated solution is one that contains more solute than the saturated solution at a given temperature.
• Polar compounds are soluble in polar solvents. Non-polar compounds are soluble in non-polar solvents.
• In endothermic process, solubility of solid solute increases with increase in temperature.
• In exothermic process, solubility of solid solute decreases with increase in temperature.

TEXTBOOK EVALUATION

Exercise 1. Choose the correct answer

1. A solution is a _____ mixture.

a. homogeneous b. heterogeneous c. homogeneous and heterogeneous d. nonhomogeneous

2. The number of components in a binary solution is _____.

a. 2 b. 3 c. 4 d. 5

3. Which of the following is the universal solvent?

a. Acetone b. Benzene c. Water d. Alcohol

4. A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called _____.

a. Saturated solution b. Unsaturated solution c. Supersaturated solution d. Dilute solution

5. Identify the nonaqueous solution.

a. sodium chloride in water b. glucose in water c. copper sulphate in water d. sulphur in carbon-di-sulphide

6. When pressure is increased at constant temperature the solubility of gases in liquid _____.

a. No change b. increases c. decreases d. no reaction

7. Solubility of NaCl in 100 ml water is 36 gram. If 25 gram of salt is dissolved in 100 ml of water how much more salt is required for saturation _____.

a. 12 gram b. 11 gram c. 16 gram d. 20 gram

8. A 25% alcohol solution means

a. 25ml alcohol in 100ml of water b. 25ml alcohol in 25ml of water c. 25ml alcohol in 75ml of water d. 75 ml alcohol in 25ml of water

9. Deliquescence is due to _____.

a. Strong affinity to water b. Less affinity to water c. Strong hatred to water d. Inertness to water

10. Which of the following is hygroscopic in nature?

a. ferric chloride b. copper sulphate pentahydrate c. silica gel d. none of the above Exercise II. Fill in the blanks

1. The component present in lesser amount, in a solution is called .

2. Example for liquid in solid type solution is .

3. Solubility is the amount of solute dissolved in g of solvent.

4. Polar compounds are soluble in solvents.

5. Volume percentage decreases with increases in temperature because .

Exercise III. Match the following

1. Blue vitriol		– CaSO₄.2H₂O
2. Gypsum		– CaO
3. Deliquescence		– CuSO₄.5H₂O
4. Hygroscopic		– NaOH

Exercise IV. True or False: (If false give the correct statement) 1. Solutions which contain three components are called binary solution. 2. In a solution the component which is present in lesser amount is called solvent. 3. Sodium chloride dissolved in water forms a non-aqueous solution. 4. The molecular formula of green vitriol is MgSO₄.7H₂O 5. When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature. Exercise V. Short answer

1. Define the term: Solution

2. What is meant by binary solution

3. Give an example each

i) gas in liquid ii) solid in liquid iii) solid in solid iv) gas in gas

4. What is aqueous and non-aqueous solution? Give an example.

5. Define Volume percentage

6. The aquatic animals live more in cold region Why?

7. Define Hydrated salt.

8. A hot saturated solution of copper sulphate forms crystals as it cools. Why?

9. Classify the following substances into deliquescent, hygroscopic:

Conc.Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride, and Gypsum salt.

Exercise VI. Long answer

1. Write notes on

i) saturated solution ii) unsaturated solution

2. Write notes on various factors affecting solubility.

3. a) What happens when MgSO₄.7H₂O is heated? Write the appropriate equation

b) Define solubility

4. In what way hygroscopic substances differ from deliquescent substances.

5. A solution is prepared by dissolving 45gram of sugar in 180 gram of water. Calculate the mass percentage of solute.

6. 3.5litres of ethanol is present in 15litres of aqueous solution of ethanol. Calculate volume percentage of ethanol solution.

Exercise VII. HOT

1. Vinu dissolves 50 gram of sugar in 250 ml of hot water, Sarath dissolves 50 gram of same sugar in 250ml of cold water. Who will get faster dissolution of sugar? and Why?

2. 'A' is a blue coloured crystaline salt. On heating, it loses blue colour and to give 'B'. When water is added, 'B' gives back to 'A'. Identify A and B, write the equation.

3. Will the cool drinks give more fizz at top of the hills or at the foot? Explain.

REFERENCE BOOKS

1. Properties Liquids Solutions John Murrell 2nd Edition.

2. Fundamental Interrelationships Between Certain Soluble Salts and Soil Colloids (Classic Reprint) Hardcover, by Leslie Theodore Sharp

📝 Summary

1. Solutions are homogeneous mixtures of two or more substances where the solute (lesser amount) dissolves uniformly in the solvent (larger amount)
2. Solutions are essential for life on Earth - sea water contains dissolved salts, air is a mixture of gases, and human body fluids like blood and urine are solutions
3. Binary solutions contain two components (one solute and one solvent) while ternary solutions contain three components (two solutes and one solvent)
4. The process of uniform distribution of solute into solvent is called dissolution, making the mixture homogeneous
5. Solutions can be classified based on the physical states of solute and solvent, with substances existing in solid, liquid, or gas phases