Board of Intermediate & Secondary Education, Rawalpindi Chemistry Chapter 6 Solutions Short questions, long questions, and MCQs Pdf download.
- Solutions MCQs chapter 6 class 9th
i) Mist is an example of solution:
- A. liquid in gas
- B. gas in liquid
- C. solid in gas
- D. gas in solid
ii) Which one of the following is a ‘liquid in solid’ solution?
- A. sugar in water
- B. butter
- C. opal
- D. fog
iii) Concentration is ratio of:
- A. solvent to solute
- B. solute to solution
- C. solvent to solution
- D. both a and b
iv) Which one of the following solutions contains more water?
- A. 2 M
- B. 1 M
- C. 0.5 M
- D. 0.25 M
v) A 5 percent (w/w) sugar solution means that:
- A. 5 g of sugar is dissolved in 90 g of water
- B. 5 g of sugar is dissolved in 100 g of water
- C. 5 g of sugar is dissolved in 105 g of water
- D. 5 g of sugar is dissolved in 95 g of water
vi) If the solute-solute forces are strong enough than those of solute-solvent forces. The solute:
- A. dissolves readily
- B. does not dissolve
- C. dissolves slowly
- D. dissolves and precipitates.
vii) Which one of the following will show negligible effect of temperature on its solubility?
- A. KCl
- B. KNO3
- C. NaNO3
- D. NaCl
viii) Which one of the following is heterogeneous mixture?
- A. milk
- B. ink
- C. milk of magnesia
- D. sugar solution
ix) Tyndall effect is shown by:
- A. sugar solution
- B. paints
- C. jelly
- D. chalk solution
x) Tyndall effect is due to:
- A. blockage of beam of light
- B. non-scattering of beam of light
- C. scattering of beam of light
- D. passing through beam of light
xi) If 10 cm3 of alcohol is dissolved in 100 g of water, it is called:
- A. % w/w
- B. %w/v
- C. % v/w
- D. %v/v
Short Questions Solutions cha 6 bise rawalpindi board
Q.1) Why suspensions and solutions do not show Tyndallthe effect while colloidAnswer:
The Tyndall effect is the scattering of a light beam because of particles. Suspension particles are quite big to be seen by the naked eye. They block the light completely and do not let it pass. Solution particles are so small that they cannot scatter the ray of light. Therefore, suspension and solution do not show the Tyndall effect.
In colloids, particles are big enough to scatter the ray of light. Thus, colloids show the Tyndall effect.
Q.2) What is the difference between solutions, colloids, and suspensions?
Answer:
The difference between solutions, colloids, and suspensions is the particles’ size. Particles of solutions are so small that they are not visible to the naked eye. Particles of suspensions are big enough to be seen with the naked eye. The particle size of colloids is intermediate between solutions and suspensions.
Q.3) Why does the suspension not form a homogeneous mixture?
Answer:
To form a homogenous mixture, particles should dissolve and form a uniform mixture. The suspension contains solid particles that are sufficiently large to be seen by the naked eye and do not dissolve. These particles settle out on standing. Therefore, no homogenous mixture can be formed. Examples of suspensions are fine sand in water, tomato juice, etc.
Q.4) How will you test whether or not the given solution is a colloidal solution?
Answer:
We will pass a ray of light through the solution. If the light passed scatters around, it means the solution is colloidal. But it is either a suspension or a true solution if it does not scatter light.
Q.5) Classify the following into true solution and colloidal solution:
Blood, starch solution, glucose solution, toothpaste, copper sulfate solution, and silver nitrate solution.
Answer:
| True solution | Colloidal solution |
| Glucose solution Copper sulphate solution Silver nitrate solution | Blood Starch solution Toothpaste |
Q.6) Why do we stir paints thoroughly before using?
Answer:
Paint is a suspension. It is a heterogeneous mixture of undissolved particles. These particles settle down on standing. Therefore, we stir paint thoroughly so the particles become suspended again in the mixture and paint can be used.
Q.7) Which of the following will scatter light and why sugar solution, soap solution, and milk of magnesia.
Answer:
Soap solution will scatter light because it is a colloidal solution. The particles in this solution cannot be seen by the naked eye but are big enough to scatter the light.
The other two solutions do not scatter light. This is because sugar solution particles are too small, and particles of milk of magnesia are so big that they block the light beam.
Q.8) What do you mean, like dissolves like? Explain with examples
Answer:
‘Like dissolve like’ states polar solutes dissolve in polar solvents, and non-polar solutes dissolve in non-polar solvents. In the case of a polar solute and non-polar solvent and vice versa, solute tends to be insoluble or only soluble to a small degree.
For example, NaCl and water both are polar. Therefore, water will dissolve NaCl. Kerosene oil is non-polar, and water is polar, so water will not dissolve kerosene oil.
Q.9) How does the nature of attractive forces of solute-solute and solvent-solvent affect the solubility?
Answer:
Solubility depends upon the relation between the solute and solvent. For a solute to dissolve in the solvent, the attractive forces of solute-solvent must overcome the attractive forces of solute-solute and solvent-solvent. If the attractive solute-solute and solvent-solvent forces are stronger than solute-solvent forces, the solute does not dissolve and remains insoluble.
Q.10) How can you explain the solute-solvent interaction to prepare a NaCl solution?
Answer:
When NaCl is added to water, the positive end of water molecules attracts the Cl- ions, and the negative end attracts Na+ ions.
These attractive forces are strong enough to overcome the attraction between Na+ and Cl- ions in NaCl crystal. These ions separate from each other. Therefore, NaCl dissolves readily in water.
Q.11) Justify with an example that salt’s solubility increases with the temperature increase.
Answer:
Some substances dissolve in water by absorbing heat. They require heat to break the attractive forces between their ions. The solubility of such solutions increases with an increase in temperature.
For example solubility of KNO3 increases with an increase in temperature.
Q.12) What do you mean by volume/volume %?
Answer:
It is the volume in cm3 of a solute dissolved in 100 cm3 of the solution. For example, 10% of alcohol solution means 10 cm3 of alcohol dissolved in sufficient water to make 100 cm3 of the solution. Calculation of this ratio is carried out by using the following formula:
Long questions
Q.1) What is a saturated solution, and how is it prepared?
Answer:
Saturated solution
The solution is a homogeneous mixture of two or more substances. The substance present in a smaller quantity in a solution is called the solute. The substance which is present in a larger quantity is called the solvent. The solute is dissolved in a solvent.
The saturated solution is one in which no more solute can be dissolved at a particular temperature. It contains as much dissolved solute as possible. Any more solute added settles as crystals at the bottom of the container.
On the particle level, the saturated solution has an undissolved solute in equilibrium with the dissolved solute.
Solute (crystallized) ⇔ Solute (dissolved)
Preparation of saturated solution
Add a small amount of solute into a solvent. It will dissolve easily. Keep the solvent at the desired temperature and add more solute; it will dissolve again. Continue to add the solute until no more of it dissolves. The solution has become saturated, and any solute added further will settle down.
A saturated solution can also be prepared by evaporating solvent from an unsaturated solution. The solvent can be evaporated by permitting air circulation or by heating the solvent.
Q.2) Differentiate between dilute and concentrated solutions with a common example.
Answer:
| Dilute solution | Concentrated solution |
| A solution that contains a relatively small amount of solute or large amount of solvent is called a dilute solution. | A solution that contains a relatively large amount of solute or small amount of solvent is called a concentrated solution. |
| Tap water is an example of a dilute solution. It contains very small quantities of dissolved minerals. | Brine is an example of concentrated solution of salt in water. |
Q.3) Explain how dilute solutions are prepared from concentrated solutions?
Answer:
Dilution is the process of making a concentrated solution less concentrated. A known volume of the stock solution is transferred to a new container and brought to a new volume. The total amount of solute remains the same before and after dilution. The dilution prepares the new solution according to the following formula:
M1V1 = M2V2
Where M1 is the concentration of the stock solution, V1 is the volume of the stock solution being diluted, M2 is the dilute solution’s concentration, and V2 is the dilute solution’s volume.
For example, we are to make 250 cm3 of 0.1 M solution of sodium hydroxide (NaOH) from a stock solution of concentrated NaOH i.e., 4.0 M. The 0.1 M solution is prepared by the dilution according to the following calculations:
M1 = 4.0 M V1 =?
M2 = 0.1 M V2 = 250 cm3
Putting the values in the above equation:
4.0 x V1 = 0.1 x 250
Take 6.25 cm3 of the solution and put in a measuring flask of 250 cm3. Add water up to the mark. Now the solution is 0.1 M solution of NaOH.
Q.4) What is molarity, and give its formula to prepare molar solution?
Answer:
Molarity
Molarity is a concentration unit. The properties and behaviour of some solutions do not depend only on the nature of the solute and solvent of a solution but also on the concentration of the solute in the solution. Chemists use many units to express concentration; however, molarity is the most common unit.
Molarity is the concentration of a solution expressed as the number of moles of solute dissolved in one dm3 of solution. For example, a 0.25 M NaOH solution contains 0.25 moles of sodium hydroxide in 1 dm3 of solution. M. represents it. The formula used for preparing molar solution is:
Preparation of molar solution
To make a one molar solution, 1 mole (molar mass) of solute is dissolved in a final volume of 1 dm3 of solution. For example, to prepare one molar solution of sodium chloride (NaCl) we need one mole of sodium chloride.
One mole of NaCl = 58.5 g of NaCl
We will dissolve 58.5 g of NaCl in sufficient water to make the total volume 1 dm3.
Q.5) Explain the solute-solvent interaction for the preparation of the solution.
Answer:
Solubility depends upon the relation between the solute and solvent. For a solute to dissolve in the solvent, the attractive forces of solute-solvent must overcome the attractive forces of solute-solute and solvent-solvent. If the attractive solute-solute and solvent-solvent forces are stronger than solute-solvent forces, the solute does not dissolve and remains insoluble.
For example, when NaCl is added to water, the positive end of water molecules attracts the Cl- ions, and the negative end of water molecules attracts Na+ ions. These attractive forces are strong enough to overcome the attraction between Na+ and Cl- ions in NaCl crystal. These ions separate from each other. This shows that the attractive forces of solute-solvent, in this case, are strong enough to overcome the attractive forces of solute-solute and solvent-solvent. Therefore, NaCl dissolves readily in water.
Q.6) What is the general principle of solubility?
Answer:
The general principle of solubility
The general principle of solubility is ‘like dissolves like. This principle states that polar solutes dissolve in polar solvents and non-polar solutes dissolve in non-polar solvents. Liquids with the same type of intermolecular forces are soluble in each other.
i) Non-polar substances have little water solubility. Hydrogen bonding holds water molecules together, while non-polar molecules are held together by London forces. For non-polar substances to dissolve in water, hydrogen bonds must be broken. But there isn’t enough energy available to break the hydrogen bonds. So, non-polar substances do not dissolve in water. For example, kerosene oil is non-polar, so water will not dissolve it.
ii) Ionic solids and polar covalent compounds dissolve in water. For example, NaCl and water both are polar. Therefore, water will dissolve NaCl. CH3OH is capable of forming hydrogen bonds. The intermolecular forces between alcohol and water molecules are similar to those in pure alcohol and water. Therefore, alcohol is soluble in water.
iii) Non-polar substances are soluble in non-polar solvents. For example, naphthalene is soluble in carbon tetrachloride.
Q.7) Discuss the effect of temperature on solubility.
Answer:
Effect of temperature on solubility
It is found that the solubility of most compounds depends strongly on temperature. Generally, it seems that solubility increases with the increase in temperature, but in some cases, the temperature hurts solubility. And in some cases, solubility does not change with an increase in temperature.
Decrease in solubility with temperature
If heat is given off during the dissolution of solid, solubility decreases with an increase in temperature. The solute-solvent interactions are stronger than the attractive forces among solute particles. Therefore, energy is released. The net dissolving reaction is exothermic. QQQQQQQQQThe addition of more heat inhibits the dissolving reaction since excess heat is already being produced.
For example solubility of Na2SO4 in water decreases with an increase in temperature.
Increase in solubility with temperature
Suppose heat is absorbed during the dissolution of solid, and solubility increases with an increase in temperature. Heat is required to break the bonds holding the solute molecules in the solid. This dissolving process is endothermic. Adding more heat provides energy to break bonds in the solid, facilitating the dissolving reaction.
For example solubility of KNO3 and KCl in water increases with an increase in temperature.
No change in solubility with temperature
In some cases, heat is neither absorbed nor given off during the dissolution of solid. In such cases, the temperature has a minimum effect on the solubility.
For example solubility of NaCl in water almost does not change with an increase in temperature.
Q.8) Give the five characteristics of colloid.
Answer:
Characteristics of colloid
i) Colloid consists of particles of the size from 1 nm to 1000 nm. The particle size of the colloid is intermediate between true solution and suspension.
ii) Colloids look homogeneous to the naked eye. However, they are heterogeneous mixtures. Colloidal particles do not settle under gravity unless subjected to high-speed centrifugation. Therefore, they are quite stable.
iii) Colloidal particles are large but cannot be seen with the naked eye. They are visible under an ultra-microscope.
iv) Colloidal particles scatter rays of light and show the Tyndall effect.
v) Colloidal particles are large but can pass through a filter paper.
Q.9) Give at least five characteristics of the suspension.
Answer:
Characteristics of suspension
(i) Suspension is a heterogeneous mixture. Particles remain undissolved.
(ii) The size of the particles in suspension is larger than 10-5 cm in diameter.
(iii) The suspended particles are big enough and can be seen with naked eyes or under a microscope.
(iv) The particles in suspension settle down if left undisturbed. Therefore, they are unstable.
(v) The solute particles in suspension are quite large and cannot pass through filter paper.
(vi) Particles block the ray of light and do not let it pass.
