Chemistry Chap 10 Solutions and Colloids Class 11 Notes KPK
Class 11 Chemistry notes according to KP Textbook Board syllabus. Contains solved exercises, review questions, MCQs, important question, 2021 notes.
Table of Contents
1st-Year Chemistry Notes Chapter 10 for kpk 2021
Q.2) How a given mixture can be differentiated into a true solution or coarse suspension.
Solution is a homogeneous mixture of two or more substances.
Suspension is a heterogeneous mixture of two or more substances.
They do not scatter light.
They scatter light.
Mostly they are transparent in appearance.
They are opaque in appearance.
Solute particles in solution are small molecules are ions and they do not settle down with time.
Solute particles in suspension are large in size and settle down with time.
When solution is filtered, the solute particles pass through the filter paper.
When suspension is filtered, the solute particles are retained on the surface of filter paper.
Salt or sugar dissolved in water is the example of true solution.
Mixture of clay and water is an example of suspension.
Q.3) Classify colloids on basis of their behaviour towards medium and the physical states of matter.
Answer: Classification of colloids Colloids can be classified on the basis of their behavior towards medium and the physical state of matter (components). Both types of classifications is discussed below.
Behavior towards medium: Depending on the forces of attraction or repulsion between the dispersed particles and dispersion medium, a colloidal suspension may be Lyophilic and Lyophobic.
If a force of attraction exists between the particles of the dispersed phase and dispersion medium, the sol (colloid) is called lyophilic. If the dispersion medium is water the sol is termed as hydrophilic. Homogenized milk is an example of a hydrophilic sol.
Lyophobic Sol If the force of repulsion exists between the dispersed phase and the dispersion medium, the sol is termed as lyophobic. In this case, if the water is used as a dispersion medium the sol is termed as hydrophobic. Lassi (casein suspended in water) is an example of hydrophobic sol.
AS2S3, Gold or Fe(OH)3 sol, solution of high polymers, Milk, Lassi.
Colorless glasses, Minerals, Alloys and mixed crystals.
Q.4 a) Explain and differentiate the following concentration units.
Answer: Concentration Units
The concentration of a solution can be expressed in many ways. Some of the concentration units are discussed below. 1) Percentage Composition: The percentage composition of the solution is the percentage amount of solute dissolved in 100 parts of the solution. It is expressed in four ways. Percentage by Weight (W/W%) It is the number of grams of solute dissolved an insufficient amount of solvent to make 100 g of solution.
2) Molarity (M): The molarity of a solution is the number of moles of solute per dm3 of solution. Molarity = Moles of solute / Volume of solution (dm3) The molarity of a solution is temperature-dependent because the volume of a solution changes with a change in temperature which changes polarity. 3) Molality (m): The molality is defined as the number of moles of solute present in 1 kg of solvent. The formula for the calculation of molality is as follows, Molality = Moles of solute / Mass of solvent (kg) Unlike molarity, the molality of a solution does not change with a change in temperature. 4) Mole Fraction: The mole fraction of any component of a solution is defined as the number of moles of that particular component divided by the total number of moles of all the components in the solution. If n1 is the number of moles of solvent and n2 is the number of moles of solute. X1 and X2 are mole fractions of solvent and solute respectively. Mathematically, X1 and X2 are expressed as, Mole fraction of solvent, X1 = n1 / n1 + n2 Mole fraction of solute, X2 = n2 / n1 + n2 Like molality, mole fraction is also independent of temperature.
Q.4 b) Calculate Molarity (M) of the following solutions.
i) 2.0 g of H2SO4 /2 dm3 of H2O ii) 0.4 g of NaOH/100 cm3 of H2O iii) 0.5 g of Na2CO3/250 cm3 of H2O
Answer: i) 2.0 g of H2SO4 /2 dm3 of H2O Solution: Mass of H2SO4 = 2.0 g Volume of Solution = 2 dm3 Calculations H = 1 g/mol | S = 32 g/mol O = 16 g/mol Molar mass of H2SO4 = 2(1.008) + 32 + 4(16) = 98 g/mol
ii) 0.4 g of NaOH/100 cm3 of H2O Solution: Mass of NaOH = 0.4 g Volume of Solution = 100 cm3 Calculations Na = 23 g/mol O = 16 g/mol H = 1 g/mol Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
iii) 0.5 g of Na2CO3/250 cm3 of H2O Solution: Mass of Na2CO3 = 0.5 g Volume of Solution = 250 cm3 Calculations Na = 23 g/mol C = 12 g/mol O = 16 g/mol Molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 106 g/mol
Q.7) Give the statement of Roult’s law. Explain the lowering of vapour pressure of a solution based on this law.
Answer: Roult’s Law:
“The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in solution”.
Lowering of Vapour Pressure: A dissolved solute lowers the vapour pressure of a liquid solvent in which it is dissolved. The lowering of vapour pressure of the solvent can be explained by the Roult’s law. Solution of component A and B is formed. A is solvent of mole fraction X1 and B is solute of mole fraction X2. Po is the vapour pressure of pure solvent and P is the vapour pressure of solution. According to Roult’s law, vapour pressure P is given by,
P = PoX1 Eq.(1) Since X1 in any solution is less than unity, P must always be less than Po. Consequently, solution of a solute in a solvent tends to a lowering of vapour pressure of the pure solvent. Furthermore, if the solute is non-volatile it does not contribute to the total vapour pressure, and hence the total vapour pressure above the solution which is due to solvent only is always less than Po. The extent of the vapour pressure lowering ΔP is, ΔP = Po – P ΔP = Po – PoX1 ΔP = Po(1 – X1) ΔP = PoX2 Eq.(2) According to Eq.(2), the lowering of the vapour pressure of solution depends both on vapour pressure of pure solvent and the mole fraction of solute in solution. In other words it depends on the nature of solvent and the concentration of solute, but not on the nature of solute. However if we consider the relative lowering of vapour pressure i.e. the ratio ΔP/Po, then from equation (2) we can write,
It shows that relative lowering of vapour pressure of the solution depends only on the mole fraction of solute and is completely independent of either the nature of solute or solvent. It is also independent of temperature.
Q.8) 0.0874 grams of ethyl alcohol (M.wt = 46) dissolved in 20.0 g of H2O, produced a depression of 0.177 K in the freezing point of solvent. Calculate the cryscopic constant of water.
Answer: Data: Weight of ethyl alcohol = W2 = 0.0874 g Weight of water = W1 = 20 g Freezing point depression = ΔTf = 0.177 K Molecular weight of ethyl alcohol = M2 = 46 g/mol Required: Cryoscopic constant of water = Kf = ? Solution: Cryoscopic constant of water can be calculated by using this formula,
Rearranging for Kf and putting the values,
Q.9) Explain the phenomenon of osmosis and the pressure exerted in this process.
“Osmosis is the spontaneous net movement of solvent molecules through a semi-permeable membrane into a region of higher solute concentration, in the direction that tends to equalize the solute concentrations on the two sides”.
Explanation: The phenomena of osmosis was first reported by Abbe Nellet in 1784. For low molecular mass solutes in water the best semipermeable membrane is a film of copper ferrocyanide Cu2[Fe(CN)6]. For high molecular weight solutes in organic solvents, the most frequently used membrane are thin films of either cellulose or cellulose nitrate.
In an experiment, a concentrated aqueous solution of sugar is taken in a thistle funnel and its mouth is tightly bound by egg membrane (acting as semipermeable membrane) and inverted in a beaker of water. The level of sugar solution is marked in the glass tube of thistle funnel. After sometime the level of sugar solution increases in the glass tube. This is due to the movement of water molecules from beaker (Higher potential) to the sugar solution (Lower potential). If pressure is applied on the glass tube, the water molecules from sugar solution move to the pure solvent. This movement of water molecules from lower concentration region (sugar solution) to higher concentration region (pure solvent) is called reverse osmosis. It consists of two steps.
Endosmosis: The flow of solvent molecules to the solution. Exosmosis: The flow of solvent molecules from solution to pure solvent. Exosmosis is also called reverse osmosis. Osmotic Pressure: Consider the figure shown below.
The closed chamber is divided into two sections A and B by a semipermeable membrane. A is fitted at one end with a moveable piston P and is filled with a solution of some solute. The chamber B is filled with solvent of the same solution taken in A. Because of osmosis the solvent will tend to pass through the membrane into the solution and displace the piston P upward. The motion of the piston and osmosis of the solvent can be prevented by the application of pressure on the piston in order to keep it in original position.
“The mechanical pressure which must be applied on a solution to prevent osmosis of the solvent into the solution through a semipermeable membrane is called the osmotic pressure of the solution”.
This pressure for the given solution depends on a number of factors but is independent of the nature of membrane. Pfeffer as a result of his experiments proved that at constant temperature, the osmotic pressure is directly proportional to the concentration of solution.
Q.10) In an experiment 1.0 C° depression of freezing point was calculated of a solution of 25.6 grams of benzene containing 6.4 g of an organic substance. Calculate M.wt of the substance (kf for C6H6 = 5.12).
Answer: Data: Depression of freezing point = ΔTf = 1.0oC Weight of benzene = W1 = 25.6 g Weight of Weight of organic substance = W2 = 6.4 g Kf for C6H6 = 5.12 Required: Molecular weight of organic substance = M2 = ? Solution: Molecular weight of the unknown organic substance can be calculated by the given formula,
Q.11) Calculate M.wt of Iodine (l2) for a solution containing 1.19 g of I2 in 35.0 grams of ether. The raise in boiling point was observed by 0.296°C (kb for ether = 2.22).
Answer: Data: Weight of Iodine (I2) = W2 = 1.19 g Weight of ether = W1 = 35 g ΔTb = 0.296oC Kb for ether = 2.22 Required: Molecular weight of Iodine = M2 = ? Solution: Molecular weight of Iodine (solute) can be calculated by the given formula,