Topic-wise Long Answers of Grade IX Chemistry Notes on Chapter "Solutions", For Higher Grades in all types of Exams.
Unlock the secrets of understanding solutions with our comprehensive grade IX chemistry notes on Long Question's answers the chapter "Solutions." Designed for higher grades, these notes provide topic-wise long answers to help you master the subject matter and excel in all types of exams. From the properties of solutions to the types of solutions, these notes cover it all, giving you a solid foundation to build upon. Don't struggle with understanding solutions any longer, read our notes and ace your exams today!
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What do you mean by Concentration units of Solution?
Concentration
The amount of solute dissolved in a given amount of solvent is called the Concentration of that solution.
A solution is said to be standard solution if its concentration is known.
To know the concentration of solution we use concentration units of solution. So concentration unit express the concentrations of solution. There are many concentration units but mostly we use following two.
I. Percentage composition of Solution
The number of parts of solute present in 100 parts of solution is called percentage composition of that solution. (OR)
The fraction of a solute in a solution multiplied by 100 is called percentage composition of that solution.
The percentage composition of a Solution can be expressed in following four concentration units.
a. Mass-mass (m/m%) percentage
It is the number of grams by mass of solute present in 100 grams by mass of a solution.
For example: If in 100g of Solution 5g is solute, then obviously 95g will be solvent and will be written as 5% m/m solution.
- % mass/mass = (Mass of solute (g)) / (Mass of solute (g)+Mass of solvent) x 100
- % mass/mass = (Mass of solute (g)) / (Mass of solution) x 100
b. Mass-Volume (m/V%) Percentage
It is the number of grams by mass of solute present in 100 cm3 (100 mL) of solution.
For example : A 5% m/V solution means 5g of solute dissolved in a solvent to make 100 cm3 of solution.
- % mass/volume = (Mass of solute (g)) / (Mass of solution in cm3 ) x 100
c. Volume-mass (V/m%) Percentage
It is the volume in cm3 of a solute dissolved in 100 g of solution.
For example : A 5% V/m Solution means 100 cm3/ml of solute dissolved in a solvent to make 100 g of Solution.
- % volume/mass = (Volume of solute (cm3 )) / (Mass of solution in (g)) x 100
d. Volume-Volume (V/V%) Percentage
It is the volume in cm3/ml of a solute dissolved per 100 cm3 of solution.
For example : A 5% V/V Solution means that 5cm3 of a solute is dissolve in 95 cm3 of solvent to make 100 cm3/ml of the solution.
- % volume/volume = (Mass of solute (cm3 )) / (Mass of solution (cm3)) x 100
II. Molarity
The number of moles of Solutes dissolved per dm3 of solution is called Molarity of that Solution.
Representation of Molarity : Molarity is represented by "M".
Formula of Molarity : Molarity (M) = No of moles of Solute / (Volume of Solution in dm3 (liter))
Or M = Amount of Solute in gram / (M.wt of Solute x Volume of Solution in dm3 )
Unit of Molarity : The unit of Molarity is gmol-1 i.e. (gram-per-mole).
Example: Calculate the molarity of a solution containing 7.50 mole of CaCO3 in enough water to make 1.50 dm3 of solution.
Solution:
As we know that M = (Number of moles of solute) / (Volume of solution in dm3)
So M = (7.50 moles) / (1.50 dm3)
or M = 5 mol.dm-3 Ans
Explain problems involving the molarity of a solution. Give examples.
Problems involving molarity of a solution
In order to find out the molarity of a solution we must know the molar mass of a solute.
Examples:
- 1-molar solution of Glucose (C6H12O6) means, one mole (180 grams) of Glucose per dm3 of solution.
- 1-molar solution of Sodium hydroxide (NaOH) contains one mole of NaOH (40 grams) in one dm3 of solution.
As the symbol for molarity is “M”, and the concentration of 1-molar solution of NaOH is written as “1M NaOH”.
Preparation of 1-molar NaOH solution
One mole of NaOH has a mass of 40g. If 40g of NaOH is dissolved in enough water to make exactly 1dm3 of solution, the solution is now 1-molar solution.
If 20 g of NaOH (which is 0.5 moles) is dissolved in enough water to make one dm3 solution, a 0.500M NaOH is produced. This relationship between molarity, moles and volume may be expressed in the following way
M = (Amount of solute in grams) / (Molecular weight of solute x Volume of solution in dm3)
M = 40 / (40 x 1 dm3) = 1 mole
M = 20 / (40 x 1 dm3) = 0.500 mole
If twice the molar mass of NaOH i.e. 80g is dissolved in enough water to make 1 dm3 of solution, a 2-molar solution is produced.
Example: 5.8g of NaCl is dissolved in water so as to make 500 cm3 solution. Determine the molarity of the solution. (At.masses: Na = 23g and Cl = 35.5 g)
Given data
Mass of solute (NaCl) in gram = 5.85 g
Volume of solution = V = 500 cm3 = 500 / 1000 dm3= 0.5 dm3
(As 1 dm3 = 1000 cm3)
Required Molarity of solution = M = ?
Solution
As we know that
Molarity = Mass of solute in grams / (Molar mass of solute x Volume of solution in dm3)…..(i)
But Molar mass of NaCl = 1 (at.mass of Na in g) + 1 (At.mass of Cl in grams)
Or Molar mass of NaCl = 1(23 g) + 1 (35.5 g)
Molar mass of NaCl = 23 g + 35.5 g
Molar mass of NaCl = 58.5 g/mol
Now put Mass of solute =5.85 g, Molar mass of solute =58.5 g and Volume of solution = 0.5 dm3 in eq (i), we have
Molarity = (5.85 g/mol) / (58.5 gx 0.5 dm3)
Molarity = 0.2 mol/ dm3
Molarity = 0.2 M Ans
Example: Calculate the molarity of 50 cm3 of solution. Containing 7.50 grams of CH3OH.
Given data
Mass of solute CH3OH in grams = 7.50 g
Volume of solution = V = 50 cm3 = 50 / 1000 dm3= 0.05 dm3
(As 1 dm3 = 1000 cm3)
Required Molarity of solution = M = ?
Solution
As we know that
Molarity = (Number of moles) / (Volume of solution in dm3)…..(I)
Molar mass of CH3OH = 1 (12 g) + 3 (1 g) + 1(16 g) + 1 (1)
Molar mass of CH3OH = 12 g + 3 g + 16 g + 1 g
Molar mass of CH3OH = 32 g/mol
and No of moles (n) = (Mass in grams) / (Molar mass)
= (7.50 g) / (32 g/mol) = 0.2344 moles
Now put Number of moles (n) = 0.2344 moles, Volume of solution = 0.05 dm3 in eq (i), we have
Molarity = (0.2344 moles) / 0.05 dm3
Or Molarity = 4.69 M Ans
Example: What volume of water must be added to 85 cm3 of 3.5 M Na2CO3 to dilute it to 0.4 M ?
Solution: Here M1 = 3.5 M, V1 = 85 cm3 M2 = 0.41 M and V2 = ?
As M1 / V1 = M2 / V2
Or V2 = (M1 x V1) / M2
So V2 = (3.5 x 85) / 0.41 = 725.60 cm3
Now Volume of solution = 725.60 – 85 = 640.60 cm3
So 640.60 cm3 volume will be needed/added to get dilution.
What factors affect the Solubility of Solution. Explain your answer.
Factors effecting Solubility
Following factors affect the solubility of a solution greatly. Their detailed explanation is given below as
I. Temperature
Generally, temperature and solubility are directly related with each other, means that increase in temperature increases the solubility of the solutions.
Explanation
In practice solubility may or may not increase with temperature, for this reason solubility depends on following two processes of thermodynamics.
- Endothermic Process
Such a process in which heat is absorbed is called Endothermic Process.
For solutes that can absorb heat, solubility increases with increase in temperature.
For example : The solubility of Sugar in water will increase with increase in temperature, because sugar absorbs heat so Endothermic Process occurs.
- Exothermic Process
Such a process in which heat is released is called Exothermic Process.
For solutes that release heat, solubility decreases with
temperature.
For example : The solubility of Calcium Oxide (CaO) in cold water is greater than its solubility in hot water, because Calcium Oxide (CaO) releases heat and so Exothermic Process occur.
More examples : The solubility of NaCl and KBr is not effected by increase or decrease in temperature.
Solubility of some solids increases up to certain temperature and then decreases on further increase in temperature. For example, Sodium Sulphate dehydrate (Na2SO4.10H2O). Its solubility increases up to 32.4 ℃ and above 32.4 ℃ its solubility decreases because it becomes anhydrous.
II. Pressure
Pressure effect solubility of solid and liquid solutes differently from solubility of Gases. Solids and Liquids are incompressible therefore their solubility is not defeated by pressure. Gases can be compressing therefore their solubility increases with increase in pressure.
For example : Carbon-di-oxide (CO2) becomes soluble in soda water because pressure is applied on it in close bottle, but when bottle is opened up it came out with effervescence (bubbles) because decrease in pressure decreases its solubility.
III. Nature of Solvent
When solute and solvent molecules are similar in structure and properties their solubility will increase.
For example : Sodium Chloride (NaCl) is more soluble in water, because NaCl is Ionic and water is Polar (both having partial or complete ions).
IV. Nature of Solute
Different Solutes have different solubilities if they are dissolved in same Solvents. So nature of Solute also affects solubility.
For example : Salt has higher solubility than sugar when they both dissolved in water.
General principle of solubility
The general principle of solubility is “like dissolves like”, it means that
- Polar solute + Polar solvent = Soluble
- Ionic solute + Polar solvent = Soluble
- Non-polar solute + Non-polar solvent = Soluble
- Non-polar solute + Polar solvent = Insoluble
Explain the concept of Solute Solvent Interaction.
Solute Solvent Interaction
In a Solution forces of attraction are present between the
ions or molecules of solute and also between solute and solvent. This
interaction is known as Solute Solvent Interaction.
Explanation
If solute and solvent of a solution are of same nature, then forces of attraction will be stronger between solute and solvent. This means that ionic or polar solute will have greater interaction for polar solvents and same is true for non-polar solute and non-polar solvents. If solvent and solute are of different nature they will not mix easily. Solute Solvent Interaction determines the extent of solubility. Solute-solvent interaction can be understood under following two headings
I. Dissolving ionic compounds in polar solvent
Consider the Solution of Salt (NaCl) and Water (H2O), here the Partial Positive Hydrogen of water will surround negative Cl-1 of Salt and partial negative Oxygen of water will surround the positive Na+1 of salt. By this way salt will dissolve in water and a solution will be form by Hydration. In this solution Na+1 and Cl-1 will be called Salveted Ions because they both are surrounded by same solvent and are also called hydrated ions because they are surround by water molecules.
II. Dissolving ionic compounds in non-polar solvent
Ionic compounds are generally not soluble in non-polar solvents such as Carbon tetrachloride (CCl4) and Benzene (C6H6). The non-polar solvent molecules do not attract the ions of the crystal strongly enough to overcome the forces holding the crystals together.
Define the term Solution Formation. State the ways by which solution can be formed.
Solution Formation
The process by which a solute is dissolved in solvent is
called Solution Formation.
Explanation
Solution formation is also called Dissolution. For majority of solution water acts as universal solvent, because it dissolves large amount of solutes. Water is called Universal Solvent for the following two characteristics and is the ways by which a solution is formed.
I. Solution formation by Hydration
That way of solution formation in which the partial
positive end of water (solvent) surrounds Anion (negative charge) and partial
negative end surrounds Cation (positive charge) of Ionic solute is called
Solution formation by Hydration.
Explanation
Consider the Solution of Salt (NaCl) and Water (H2O), here the Partial positive Hydrogen of water will surround negative Cl-1 of Salt and partial negative Oxygen of water will surround the positive Na+1 of salt. By this way salt will dissolve in water and a solution will be form by Hydration. In this solution Na+1 and Cl-1 will be called Salveted Ions because they both are surrounded by same solvent and are also called hydrated ions because they are surround by water molecules.
The ions of solute which are surrounded by the same solvent
are called Salveted ions.
II. Solution formation by Hydrogen Bonding
That way of Solution formation by which a non-Ionic compound is dissolved in water because of Hydrogen bonding between them is called Solution formation by Hydrogen Bonding.
Explanation
Certain compounds like Ethyl Alcohol and Sugar etc which are not Ionic in nature still dissolve in water because of Hydrogen bonding between solvent and solute.
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