Study Portal 14 - Chapter 14 - The-Announcements.Com

Study Portal 14 – Chapter 14



Q1.      Fill in the blanks with appropriate words.

  1. i) A solution is the Homogeneous type of the mixture.
  2. ii) Strength of a solution in mole/liter is called Molarity.

iii)     The Solute- Solvent interaction in dissolution process is termed as Solvolysis.

  1. iv) Generally heating Increase the solution formation.
  2. v) Milk of magnesia is the example of Suspensions.
  3. vi) The Solute is that part of solution which is Smaller in quantity.

vii)    In Saturated solution no more quantity of solute can dissolve at normal temperature.

viii)   In Dissolution process, a general rule of solubility is that like is dissolves by like.

  1. ix) PPM Stands for Parts per Million.
  2. x) The solubility of gases decreases with increase in temperature.

Q2.      Choose the Correct Answers.

  1. i) The Solvent is the part of solution which is:

(a) More in quantity         (b) Less in quantity

(c) Equal in quantity          (d) Does not depend on quantity

  1. ii) The heterogeneous mixture where the particles size of solute is larger the 1200mm is called.

(a) Solutions          (b) Colloids   (c) Buffers    (d) Emulsion

iii)     In dissolving table salt , heating accelerates the Solvolysis:

(a) True                                        (b) False

(c) Both statements are false          (d) Both Statements are true

  1. iv) The air is a solution:

(a) True                                        (b) False

(c) Both statements are false          (d) Both statements are true

  1. v) Steel is the type of solution having.

(a)  Liquid – Solid Composition

(b) Liquid – Liquid Composition

(c) Solid- Solid Composition

(d) Solid gas composition


Q3.      Answer the following question in short.

  1. i) What is difference between unsaturated, saturated and super saturated solution.

Ans.  A solution containing maximum amount of solute at a given temperature is called saturated.

A solution which contains lesser amount of solute than that which is required to the saturate it at a given temperature is called unsaturated solution.

The solution that is more concentrated than a saturated solution is known as super saturated solution.

  1. ii) Draw a picture to show that what happens when an ionic substance is added to water to dissolve?

Ans.  Solution formation depends upon the relative strength of attractive forces between solute- solvent, for example when NaCl is added in water it dissolve readily because the attractive interaction between the ions of NaCl and polar molecules of water are strong enough to overcome the attractive forces between Na+ and Cl ion sand the negative and of water dipole is oriented towards the Na+ ions

iii)     Why NaCl is dissolved easily in cold water but not in heated cooking oil.

Ans.  Solubility is related with factors, of which the nature of solute and solvent are very important. A general phenomenon which is observed during solubility is that like is dissolved by the like thus NaCl is soluble in water (both are polar) and boiling oil is non polar solvent.

  1. iv) What type’s of solution can form? Example of each types?

Ans.  Each Solution consists of two components, solute and solvent. The solute as well as solvent exists as gas liquid or solid so, depending upon the nature of solute and solvent difference types of solutions may form.

S# Solute Solvent Example of Solution
1 Gas Gas Air, Mixture of H2 and He in weather balloons

Mixture of N2 and O2  in cylinders for respirations

2 Gas Liquid Oxygen in water , carbon Dioxide in water
3 Gas Solid Hydrogen absorbed on palladium
4 Liquid Gas Mist, fog liquid air pollutants
5 Liquid Liquid Alcohol in water , Benzene and toluene
6 Liquid Solid Butter , Cheese
7 Solid Gas Dust Particles or smoke in air
8 Solid Liquid Sugar in water
9 Solid Solid Metal alloys (Brass, Bronze) opals
  1. v) What is meant by molarity?

Ans.  Molarity is concentration unit defined as number of solute dissolved in one dm3 of  Solution it is represented by M

Molarity (M) = Number of Moles of solute / Volume of solution in liter

  1. vi) Explain dilution of solution with an example.

Ans.     Dilute molar solution is prepared from a concentrated solution of known molerity as suplained below suppose we are to make 100 cm3 of 0.01M solution from given 0.1 M Solution of Potassium permanganate  First 0.1M Solution is prepared by dissolving solution . Then 0.01 M solution is prepared by the dilution according to following

Concentrated solution                                Dilute Solution

M1V1   = M2V2

Where        M1        = 0.1 M

V1        = ?

V2        = 100cm3

M2       = 0.01 M

Putting the values in above equation we got

V1 ´ 0.1        = 0.01 ´ 100

V1                 = 0.01 ´ 100/0.1

V1                 = 10 cm3

Concentrated Solution of KMNo4 has dense purple color

vii)          Explain the term Suspension?

Ans.  Suspension is a Heterogeneous mixture of un-dissolved particles in a given medium particles are big enough to be seen with naked eye. Examples are chalk in water (milky suspension) paints and milk of magnesia.

viii)         What are colloids?

Ans.  Sometime a solution consists of bigger solute particles , The particles although are evenly distributed in the solvent and appear to from a homogenous mixture but over a time they separate from the solvent system and form a heterogeneous mixture rather then homogenous phase due to having a bigger solute size such solutions are termed as the colloids.

  1. ix) Compare advantages & disadvantages of percentage & molarity over each other.

Ans.  The Percentage calculation of solution does not require the chemical formula of solute; it requires only mass or volume of solutes. Since in molarity calculation we require no of moles of solute dissolve per dm3 of solution so in order to calculate the molar mass of state, we needed a formula of solute.

  1. x) What would happen if solubility of oxygen in water increases one thousand times?

Ans.  Most aquatic organisms, just like land dwelling animals, need to “breathe” oxygen to survive.  Organisms must have a minimum amount of oxygen to survive.  Unlike an increase in water temperature which can stress or even kill aquatic organisms, an increase in dissolved oxygen is not harmful.  Although, there is a minimum amount of dissolved oxygen required in water for organisms to survive, there is no maximum amount or upper limit for dissolved oxygen.  Water can have too many hydrogen ions, nitrates, or heavy metals dissolved in it, but, in terms of dissolved oxygen, “you can’t have too much of a good thing”. Due to increase in solubility of oxygen in water, Rusting of iron will increase, will cause acid rain and oxidation will occur.


Q.4.     Answer the following question with reasoning.

  1. i) Which if the following expected to dissolve in H2O and why? (a) Alcohol (b). Hydrocarbon (c). NaCl (d) iron.

Ans.  Alcohol and NaCl are expected to dissolve in water because general phenomenon of solubility is that like dissolved by the like, the NaCl and alcohol soluble in water (they are polar)

  1. ii) Explain why some liquids are miscible in one another while other liquids are immiscible.

Ans.  Similar Solvent dissolve similar solute i.e. if the chemical structure and the electrical properties such as dipole moment of solute and solvent are similar , the solubility will increase , if these properties are different, the solute either will not dissolve or there will be very litter solubility  for example water (liquid) is polar compound and is a good solvent for polar substance such as C2H5OH (ethyl alcohol) Liquid are miscible in one another while water(liquid) cannot dissolve (immiscible) non-polar substance such as kerosene oil (liquid).

iii)     Why it is necessary to bubble air through an aquarium?

Ans.  we have observed in home aquarium , that the fish shows sings of stress on hot day , this is because of less oxygen from dissolves in the warm water due to increase of temperature so it is necessary to bubble are through an aquarium.

  1. iv) Why solubility of gases decreases with an increase in temperature?

Ans.  The Solubility of gases e.g. air in water decreases with increasing temperature because when water is heated, we will see small bubbles form at the side of the beaker before the water boils these bubbles are composed of air and come out of water in the form of bubbles. This means that solubility of air (gas) in water decrease with increasing temperature.

  1. v) Explain how can you obtain pure solute from a solution by adding more solute to the solution?

Ans.  A solution which contains amount of a solute in a particular amount of solvent, then the saturated solution is called super saturated solution. Some of the solute remains undissolved and becomes visible as a separate phase. Let us take the example of sodium Thiosulphate (Na2S2O3) in water. Prepare the saturated solution of it in water. Now add more amount of Na2S2O3 and go on heating the solution. More and more quantity of it will dissolve. Now cool the solution, when it comes to room temperature some of the Na2S2O3 will settle down. Than filter the solution and get pure solute by adding more solute.

Q5.      Define the term solution? How the solutions form?

  • A Solution is a homogeneous mixture of two or more substances OR
  • A homogeneous mixture of two or more different chemical substance with uniform chemical and physical properties is called solution.
  • A solution which consists of the only substance is called binary solution.
  • The word aqueous is derived from the Latin word aqua which means water. A solution in which water is used as a solvent is called aqueous solution.

A Solution consists of two parts , the solute and the solvent the solute is that part which is lesser in quantity while the solvent that which is larger in quantity for example  in the aqueous solution of sugar in water the sugar is solute and the water is solvent .


Q.6.Discuss Various Concentration units?

Ans.  Concentration is the proportion of a solute in a solution. it is also a ratio of amount of solute to the amount of solution or ratio of amount of solute to amount of solvent. Keep in mind that concentration does not depend upon the total volume or total amount of the solution; there are various types of units used to express concentration of solution. A few of these units are discussed here.

(a)           Percentage:- Percentage unit of Concentration refers to the Percentage to the solute present in a Solutions. A percentage of Solute can be expressed by mass or by volume. it can be expressed in terms of percentage composition by four different ways .

(b)           Percentage-Mass/Mass(% mass mass):- It is the number of grams of solute in 100 grams of solution for example 10% m/m sugar solution means . That 10g of sugar is dissolved in 90 gram of water to make 100 grams of solution. Calculation of this ratio is carried out by using the following formula.

% by mass = Mass of Solute (g) / Mass of Solution (g) ´ 100

(c)           Percentage Mass/ Volume:- It is the number of grams of Solute dissolve in 100 cm3 of solution. For example 10% m/v sugar solution 10 gram of sugar in 100 cm3 of solution the exact volume of solvent is not mentioned or it is not known.

% M/V = Mass of Solute (g) / Volume of Solution cm3 ´ 100

(d)           Percentage – Volume / Mass:- It is the volume in cm3 of a solute dissolve in 100 gram of solution . For example 10 % v/m alcohol solution in water means 10 cm3 of alcohol is dissolve in (unknown).  Volume of water so that the total weight of solution is 100 grams. In such solution the mass of solution is under consideration, total volume of the solution is not considered.

% v/m = volume of solute (cm3) /mass of solution (g) x 100

(c)           Percentage Volume / Volume:- It is the volume in cm3 of a solute per 100 cm3 of the solution for example 30 % alcohol solution means 30 cm3 of alcohol dissolve  in sufficient amount of water. So that the total volume of solution becomes 100 cm3.

% by volume=volume of solute cm3/volume of solution cm3´100


Q7.   Calculate the molarity of a solution of NaHCO3 that contains 50gm of NaHCO3 in 250 ml of water?


Mass of solute (NaHCO3)                       =          50gm

Molecular mass of solute (NaHCO3)       =          23 + 1 + 12 + 3(16)

=          23 + 1 + 12 + 48

=          84

Amount of solution in liter                     =          250ml = 0.25liter

Molarity = mass of solute/molecular mass´amount of solution

= 50 / (84 ´ 0.25)         = 50 / 21          = 2.38M

Q8.   Find out the amount of glucose required to prepare its 0.5M solution in 500ml of water?


Mass of solute (glucose) C6H12O6           =?

Molarity of solution                                = 0.5M

Molecular mass of solute (C6H12O6)       = 6 ´ 12 + 12 ´ 1 + 6 ´ 6

= 72 + 12 + 96  = 180

Amount of solution                                = 500ml or 0.5 liter

Molarity = mass of solute / (molecular mass ´ amount of


Mass of solute    = molarity        ´ molecular mass

´ amount of solution in liter

= 0.5 ´ 180 ´ 0.5         = 45 gm

Q.9.  How the temperature affects the solubility process? Give example to prove your statement.

Ans.  Temperature has major effect on the solubility of most of the substances.  Generally it seems that solubility increase with the increase of temperature, but it is not always true. When a solution is formed by adding a salt in solvent there are different possibilities with reference to effect of temperature on solubility. These possibilities are discussed here.

  • Heat is absorbed. When salts like KNO3, KCL and NaNo3 are added in water, the test tube becomes cold. It means during dissolution of these salts heat is absorbed. Such dissolving process is called Endothermic.
  • Heat is given out. On the other hand when the salts like Li2SO4 and Ce2 (So4)3 are dissolved in water, the test tube becomes warm that is heat is released his dissolution. In such cases the solubility of salt decreases with the increase of temperate.
  • No Change in heat. In some cases during a dissolution process neither the heat is absorbed nor released.  when salt like NaCl is added in water , the solution temperature remains the almost same  in such case the temperature has a minimum effect on solubility.

Q.10.   How would you differentiate between the solution, suspension and colloids?

S.No Solution Colloid Suspension
1 The particles exist in their simplest form i.e. as molecules or ions. Their diameter is 10-8cm  The particles are large consisting of many atoms, ions or molecules. The particles are of largest size. They are larger than 10-5 cm in diameter.
2 Particles dissolve uniformly through out and form a homogeneous mixture. A colloid appears to be a homogeneous but actually it is a heterogeneous mixture. Hence they are not true solution particles do not settle down for a long time. Therefore colloids are quite stable. Particles remain undissolved and form a heterogeneous mixture. Particles settle down after sometime.
3 Particles are so small that they can’t be seen with naked eye. Particles are large but can’t be seen with naked eye. Particles are big enough to be seen with naked eye.
4 Solute particles can pass easily through a filter paper. Although particles are big but they can pass through a filter paper. Solute particles can not pass through filter paper.
5 Particles are so small that they can not scatter the rays of light thus do not show tyndall effect Particles scatter the path of light rays thus emitting the beam of light i.e. exhibit the tyndall effect Particles are so big that light is blocked and difficult to pass.

Example No. 6.1:- Found out the % of a solution that has 4g of glucose dissolved in 500g of water.

Here 4g of glucose has been dissolved in 500mg of water, so mass / mass relationship exists.

The % of solution can be determined by applying formula

%         =          (mass of glucose / mass of solution) ´ 100

=          4 / 500 x 100   =          0.8%

 Example No. 6.2:- Calculate percentage of NaCl in a solution which has been prepared by dissolving 8g of NaCl in 250ml of water.

In this example the mass / volume relation exists so applying the formula.

%            = (mass of solute / volume of solution) ´ 100

%            = 8g / 250 ml ´ 100     = 3.2%

Example No.6.3:- In a face cream recipe 6ml of almond oil has been mixed in petroleum jelly to get 50g of total mixture solution. Find out the percentage of almond oil in this cream.

The relation here is volume / mass, so we will use the following formula.

Percentage       = (volume of solute / mass of solution) ´ 100

= (6ml / 50g) ´ 100      = 12%

Example No.6.4:- Calculate amount of NaOH required preparing in 0.1M solution in 500ml of water.

Molarity           = mass of solute / (molecular mass ´ amount of solution in


NaOH              = 23 + 16 + 1   = 40

Mass of solute = molarity x molecular mass x amount of solution

= 0.1 ´ 40 x 0.5

= 2 gms

Example No.6.5:- What is the volume of HCl required to prepare 500ml of 0.1M solution from the commercially provided 36M HCl acid?

This process requires the dilution process. The formula that will work is:

M1V1    = M2V2

M1        = Molarity of commercial HCl = 36M

V1        = Volume of commercial HCl  = ?

M2        = Molarity of required HCl      = 0.1M

V2        = Volume of required HCl       = 500ml

M1V1    =M2V2

V1        = M2V2 / M1

V1        = 0.1 ´ 500 / 36           = 1.39ml

Activity No.6.1:- For the preparation of vinegar 30ml of acetic acid has been dissolved in water to make the total volume of mixture one liter. Calculate the percentage of CH3COOH in this vinegar mixture.

In such type of solution the amount of both the solute and the solution are expressed in volume unit.

%         = [volume of solute (cm3) / volume of solution (cm3)] ´ 100

%         = (30 / 1000) ´ 100

%         = 3%

Activity No.6.2:- How will you prepare a 250 ml of a solution of 0.5 M NaHCO3?

Molarity           = 0.5M

Molecular mass of NaHCO3    = 23 + 1 + 12 + 3 ´ 16

= 23 + 1 + 12 + 48

= 84

Amount of solution                 =250ml  = 0.25litre

Mass of solute =?

Mass of solute             = molarity ´ molecular mass ´ amount of solution

= 0.5 x 84 x 0.25

= 10.5g