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Chapter 6 Solutions 9 class Chemistry Notes
Table of Contents
Q.1) Is seawater a solution? How would you prove with a simple experiment whether it is pure water or a solution?
Answer: Seawater is not pure and is the solution of different substances in water. It can be proved by boiling a sample of seawater. When it is boiled, the salts dissolved are left behind as residue showing that the seawater is not pure water.
Q.2) A bottle in a drug store contains a label “3 percent hydrogen peroxide.” What does it mean?
Answer: It means that 3 cm3 of hydrogen peroxide is dissolved in sufficient volume of solvent , so that the total volume of solution is 100 cm3. 3% of the bottle’s volume is pure hydrogen peroxide, whereas the remaining 97% is solvent i.e water.
Q.3) Classify the following as a solution, a colloid or suspension and explain why:
i. Milk ii. Hot cup of tea iii. Orange juice with pulp iv. Mayonnaise v. Listerine mouthwash vi. Milk of Magnesia vii. Cheese viii. Mist ix. Bottled water
Hot cup of tea Listerine mouthwash Bottled water
Milk Mayonnaise Cheese Mist
Orange juice with pulp Milk of magnesia
The components in above examples are very small and cannot be seen with naked eye.
The particles cannot be seen but they scatter a beam of light.
The particles are large and can be seen with naked eye.
Q.4) Why we stir paints thoroughly before using it?
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 mixture and paint can be used.
Q.5) Why suspensions and solutions do not show Tyndall Effect, while colloids do?
Answer: Tyndall effect is the scattering of beam of light because of particles present. Suspension particles are quite big to be seen by 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 Tyndall effect. In colloids, particles are big enough and dispersed in transparent medium to scatter the ray of light. Thus, colloids show Tyndall effect.
Long Questions 9th class chemistry notes
Q.1) Define solution? Explain types of solution on the basis of states of matter.
Answer: Solution A solution is a homogeneous mixture of two or more substances. In a solution, the substance that is present in lesser amount is called solute and the substance that is present in a larger amount is called solvent. For example, 10 g of sugar in 100 mL of water is a sugar solution in which sugar is a solute and water is a solvent. Types of solution 1. Gaseous solutions (i) Gas in gas Solute and solvent both are gases. For example, oxygen in nitrogen in air. (ii) Gas in liquid Solute is gas and solvent is liquid. For example, carbon dioxide in water in carbonated drinks. (iii) Gas in solid Solute is gas and solvent is solid. For example, hydrogen gas absorbed in palladium 2. Liquid solutions (i) Liquid in gas Solute is liquid and solvent is gas. For example, fog, humidity in air. (ii) Liquid in liquid Solute and solvent both are liquids. For example, alcohol in water. (iii) Liquid in solid Solute is liquid and solvent is solid. For example, mercury amalgams. 3. Solid solutions (i) Solid in gas Solute is solid and solvent is gas. For example, carbon particles in air in smoke. (ii) Solid in liquid Solute is solid and solvent is liquid. For example, seawater. (iii) Solid in solid Solute and solvent both are solids. For example, alloy of copper and tin.
Answer: Solubility Solubility is the number of grams of a substance (solute) dissolved in 100 g of a solvent to prepare a saturated solution at a given temperature. Different substances have different solubilities in the same amount of solvent at a specific temperature. For example, sodium nitrate is more soluble than silver nitrate in water.
Q.2 b) Explain the factors that are responsible for the solubility of a substance.
Answer: Factors affecting solubility (i) Temperature If heat is given off during dissolution of solid, solubility decreases with increase in temperature. For example solubility of Na2SO4 in water decreases with increase in temperature. If heat is absorbed during dissolution of solid, solubility increases with increase in temperature. For example solubility of KNO3 and KCl in water increases with increase in temperature. In some cases, heat is neither absorbed nor given off during dissolution of solid. In such cases temperature has minimum effect on the solubility. For example solubility of NaCl in water does not change with increase in temperature. (ii) Pressure The solubilities of solids and liquids are not affected by changing pressure because they are incompressible. The effect of pressure is observed only in the case of gases. An increase in pressure increases the solubility of a gas in a liquid. For example carbon dioxide is filled in cold drink bottles under pressure. When this bottle is opened, the pressure decreases and carbon dioxide comes out resulting in decrease in the solubility.(iii) Nature of solute and solvent (a) Non-polar substances have little water solubility. So, non-polar substances do not dissolve in water. For example, kerosene oil is non-polar, so water will not dissolve it. (b) Ionic solids and polar covalent compounds dissolve in water. For example, NaCl and water both are polar, therefore, water will dissolve NaCl. (c) Non-polar substances are soluble in non-polar solvents. For example naphthalene is soluble in carbon tetrachloride.
Q.3 a) What is the difference between a concentrated and a dilute solution? Give examples of each.
A solution that contains a relatively small amount of solute or a large amount of solvent is called a dilute solution.
A solution that contains a relatively large amount of solute or a small amount of solvent is called a concentrated solution.
Examples A solution in which 5g of sugar is dissolved in 100 cm3 of water is a dilute solution. Tap water is an example of a dilute solution. It contains very small quantities of dissolved minerals.
Examples A solution in which 8g of sugar is dissolved in 100 cm3 of water is a concentrated solution. Brine is an example of concentrated solution of salt in water.
Q.3 b) Differentiate between unsaturated, saturated and supersaturated solutions.
Answer: Difference between unsaturated, saturated and supersaturated solutions:
A saturated solution contains the maximum amount of solute at a particular temperature and is unable to dissolve the further amount of solute in it.
An unsaturated solution can dissolve the further amount of solute at a particular temperature.
A supersaturated solution contains more of the solute than is contained in the saturated solution.
Example: Add sugar in water. Not all the sugar crystals dissolve and a few settle down
Example: Add sugar in water. All the sugar crystals dissolve and none settle down
Example: To dissolve the settled sugar heat the water.
Q 4) Describe one way to prove that a mixture of a sugar and water is a solution and that mixture of sand and water is not a solution.
Answer: Sugar in water forms a homogeneous solution. Whereas, sand in water does not dissolve to form a solution. It settles down. The statements can be proved by performing filtration on both samples. When a mixture of sugar and water is filtered, all the components easily pass through the filter paper and no residue is left behind showing that the mixture is a solution. When a mixture of sand and water is filtered, sand particles are left behind as residue showing that it is a mixture and not a solution. This is because sand does not dissolve in water and is large enough to not pass through the filter paper.
Q.5) Explain the following concentration units.
a. Percentage composition
Answer: a. Percentage composition Percent composition refers to the percentage of each element present in a compound. Percentage composition can be expressed by mass or by volume. For example, one mole of water is 18 g. There are 2 moles of hydrogen atoms, so 2 x 1 = 2 g of hydrogen is present in one mole of water. There is one mole of oxygen atom, so 16 g of oxygen is present in one mole of water. To get the percentage of hydrogen: % mass of hydrogen = mass of hydrogen atoms / mass of all atoms in compound x 100 = (2 / 16 + 2) x 100 = 11% b. Molarity 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. It is represented by M. The formula used for preparing molar solution is: Molarity (M) = moles of solute / volume of solution (dm3) As, Mole = mass of solute (g) / molar mass of solute (gmol-1) So, Molarity (M) = mass of solute (g) / (molar mass of solute x volume of solution)
Numericals 9th Chemistry Chapter 6
Q.1) What is the molarity of a solution composed of 5.85 g of potassium iodide (Kl) dissolved in enough water to make 0.125 dm3 of solution?
Answer: Given that Mass of KI = 5.85 g Volume of solution = 0.125 dm3 Molar mass of KI = 39 + 127 = 166 Molarity =? Formula applied Molarity = Mass of KI / (Molar mass of KI x Volume of solution) = 5.85 / (166 x 0.125) = 0.281 M
Q.2) How many moles of H2SO4 are present in 0.500 dm3 of 0.150 M H2S04 solution?
Answer: Given that Volume of solution = 0.500 dm3 Molarity = 0.150 M Moles of H2SO4 =? Formula applied Molarity = Moles of H2SO4 / Volume of solution Moles of H2SO4 = Molarity x Volume of solution = 0.150 x 0.500 = 0.075 moles
Q.3) Suppose you wanted to dissolve 40.0g NaOH in enough H2O to make 6.00 dm3 of solution.
a. What is the molar mass of NaOH? b. What is the molarity of this solution?
Answer: Given that Mass of NaOH = 40.0 g Volume of solution = 6.00 dm3 Solution a. Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol b. Molarity = Mass of NaOH / (Molar mass of NaOH x Volume of solution) = 40 / (40 x 6) = 0.166 M
Q.4) What is the molarity of a solution of 14.0 g NH4Br in enough H2O to make 150 cm3 of solution?
Answer: Given that Mass of NH4Br = 14.0 g Volume of solution = 150 cm3 = 150 / 1000 = 0.15 dm3 Solution Molar mass of NH4Br = 14 + 4(1) + 80 = 98 g/mol Molarity =? Formula applied Molarity = Mass of NH4Br / (Molar mass of NH4Br x Volume of solution) = 14.0 / (98 x 0.15) = 0.9523 M
Q.5) Suppose you want to produce 1.00 dm3 of 3.50 M solution of H2SO4.
a. What is the solute? b. What is the solvent? c. How many grams of solute are needed to make this solution?
Answer: a. The solute is H2SO4. b. The solvent is water. c. Given that Molarity = 3.50 M Volume of solution = 1.00 dm3 Molar mass of H2SO4 = 2(1) + 32 + 4(16) = 98 g/mol Mass of H2SO4 =? Formula applied Molarity = Mass of H2SO4 / (Molar mass of H2SO4 x Volume of solution) Mass of H2SO4 = Molarity x Molar mass of H2SO4 x Volume of solution = 3.50 x 98 x 1.00 = 343 g