Comp_Chemistry_Vol_I_XII_CBSE_2020

w Types of Solutions We have studied in the lower classes that substances have been classified into two types. These are pure substances and mixtures. Elements and compounds comprise w Expressing Concentration of Solutions pure substances. With the exception of noble gases, elements donot exist independently. w Solubility However, compounds formed chemically by the combination of two or more elements w Review Problems exist independently. Mixtures fall under the category of impure substances, since w Vapour Pressure of Pure Liquid these are formed by mixing elements or compounds by physical means. Moreover, w Vapour Pressure of Liquid Solutions their composition can also vary. Mixtures have been further classified as homogeneous w Ideal and Non–Ideal Liquid Solutions and heterogeneous. Homogeneous mixtures are regarded as solutions from the point of w Constant Boiling Liquid Mixtures (Azeotropes) view of a chemist. A solution may be defined as w Review Problems w Colligative Properties and Determination of Molar a homogeneous mixture of two or more pure non–reacting substances whose composition can be varied within certain limits. Mass w Relative Lowering in Vapour Pressure Sometimes, a solution is also termed as a true solution. Please note that w Elevation in Boiling Point Temperature heterogeneous mixtures donot form a true solution. They may lead to either colloidal w Depression in Freezing Point Temperature solution or suspension. In the present unit, we will focus only on homogeneous w Osmosis and Osmotic Pressure mixtures or solutions and will discuss colloidal solutions and suspensions in Unit-4 w Review Problems on Surface Cemistry. We shall get an opportunity to learn characteristics of the w Abnormal Molecular Masses solutions such as concentration, factors influencing the solubilities of solutes in w Review Problems solvents, vapour pressure of solution, ideal and non-ideal solutions. Apart from w Concept Review these, we shall also try to understand the different types of colligative properties FOCUS—N.C.E.R.T. associated with dilute solutions. m N.C.E.R.T. In–Text Questions Sec. 1. Types of Solutions m N.C.E.R.T. Exercise m N.C.E.R.T. Exemplar Problems with Solution A solution may contain two or more substances also called components or FOCUS—BOARD EXAMINATIONS constituents. A solution which has two components is known as binary solution (e.g. m Passage Based Questions a solution of NaCl in water) while a solution with three components is called m One Word Answer Questions ternary solution (e.g. a solution of NaCl and KCl in water). Similarly even more m Multiple Choice Questions than three components may also be present. However, we shall concentrate only on m True and False Questions the binary solutions in which the component or constituent present in smaller m Assertion-Reason Questions proportion or amount is called solute while the one present in excess is known as m Questions from Board Examinations solvent. For example, in a binary solution of sugar in water, sugar acts as the m H.O.T.S. Questions solute while water is the solvent. It may be noted that out of the three states of a m H.O.T.S. Numerical Problems substance any one can act as the solute or solvent resulting in nine different types of m Assignment solutions which are listed in Table 1. m Solution to Problems for Practice FOCUS—COMPETITIVE EXAMINATIONS Binary solutions are further classified as solid, liquid and gaseous solutions m Useful Information for Competitive depending upon the nature of the component acting as solvent. Examinations 1.1 m N.E.E.T. & Other Medical Entrance Examinations m JEE Main & Other Engineering Entrance Examinations m JEE (Joint Entrance Examination) Advanced



















































SOLUTIONS 1.27 (iv) At high altitudes, the partial pressure of oxygen is less than its pressure at ground level. This means that lesser quantity of oxygen will be inhaled. This will lead to low concentration of oxygen in the blood and tissues of the people living under the prevailing conditions. They will be faced with breathing problems and this will lead to anoxia. It affects brain as well and leads to loss of memory. It is because of this reason that the climbers which have to climb to high altitudes generally carry cylinders of oxygen alongwith them in order to supplement the deficiency of oxygen. (v) The aquatic species (fish etc.) are more comfortable in cold water than in warm water. We know that the solubility of any gas in water decreases with the rise in temperature. This means that the solubility of oxygen in water is more in cold water than in warm or hot water. Since oxygen is essential for breathing, aquatic species can breathe more comfortably in cold water than in warm water due to higher percentage of oxygen present in cold water. Hence, they feel more comfortable in cold water than in warm water. Limitations of Henry’s Law Although Henry’s Law has vital applications, but it does suffer from some limitations as well. These are listed : (i) The law is applicable to gases only under ideal conditions i.e., under low pressure and at high temperature. (ii) The molecules of the gas should not dissociate or associate in solution i.e., the gas has same molecular state in the liquid as well as gaseous phases. (iii) The gas should not react chemically with the solvent to form any compound. For example, NH3(g) combines with water to form NH4OH(aq) which dissociates as follows : NH3(g) + H2O(l) ¾® NH4OH(aq) NH4+(aq) + OH–(aq) Similarly, hydrogen chloride gas or HCl (g) dissolves in water to form hydrochloric acid HCl (aq). The acid dissociates into ions as follows : HCl(g) + aq ¾® HCl (aq) H+(aq) + Cl– (aq) 122345627589 7 35723 4 According to Henry’s Law, pA = kHxA pA = Partial pressure of the gas ‘A’ over the surface of solution where xA = Mole fraction of the gas kH = Henry’s law constant Example 35 The Henry’s Law constant for the solubility ofrfoNm2 gas in water at 298K is 1·0 × 105 atm. The mole fraction of ?N2 in air is 0·8. What is the number of moles of N2 air dissolved in 10 moles of water at 298 K and 5 atm pressure (I.I.T. 2009) Solution. According to Henry’s Law, pN2 1 KHxN2 ...(i) pN2 1 xN2 2 PTotal = 0·8 × 5 atm = 4·0 atm ( xN2 is mole fraction in air) pN2 ( 4.0 atm ) ...(ii) KH (1.0 2 105 atm ) ( xN2 is mole fraction in water) From eqn. (i), xN2 1 1 1 4 2 1035 Now, nN2 1 xN2 1 4 2 1035 nN2 1 nH2O nN2 = 4 × 10–5 (nN2 can be neglected) nH2O \\ nN2 = 4 × 10–5 × nH2O = 4 ×10–5 × (10 mol) = 4 × 10–4 mol. Example 36 IAfssNu2mgeatshaist bubbled through water at 2o9f 30.K98,7hboawr.mGaivneynmthilaltimKoHlefsoroNf 2Na2t gas would dissolve in one litre of water. N2 exerts a partial pressure 293 K is 76.48 k bar. Example) (N.C.E.R.T. Solved Solution. Mole fraction of N2 ( xN2) = Partial pressure of N2 KH for N2 (0.987 bar) = (0.987 bar) = (76480 bar) = 1.29 × 10–5 (76.48 k bar)