 Good morning myself, Dr. Nitin Gramopadhyay, Department of Basic Sciences and Humanities, Walton Institute of Technology, SolarPur. Today we are going to learn solution chemistry topic and from that topic particularly we will learn the problem solving section of that topic. This is the learning outcome for today's session. At the end of this session students will be able to describe the terms such as normality, molarity and molality. Also by learning these terms we will apply the definition part of these terms and we will learn how to calculate normality, molarity and molality of a given solution or to calculate what is the weight of solute required to prepare a given solution. So these are the two outcomes of today's session. In one part we are learning the theoretical part of this concept and in second part we are applying the concept of these terms in calculating the normality, molarity and molality of a given solution or what is the weight of solute required to prepare a given solution. This is the content we are going to learn in this session. First we will see what is meant by solution chemistry that is the introduction for solution chemistry, various terms used in solution chemistry and then calculation of strength of solution. Analytical chemistry is one of the important branch of chemistry which mainly deals with analysis of given material. Analysis of a given material can be either done by instrumental method or non-instrumental method. In both methods various solutions are reacted with the material for its analysis and the results of analysis mainly depends upon how correctly all solutions are prepared. Similarly solution strength is measured in the form of its normality, molarity or molality and the solution can be prepared either by dissolving solute in water which is known as aqueous solution or the solute is dissolved in suitable solvents then it is known as non-aqueous solution. So in short what do you mean by solution chemistry? Solution chemistry is the main part of the analytical chemistry where we are analyzing the given sample of material and how the analysis is carried out? The analysis is carried out by treating the material with a suitable solution and then the observations are recorded and then the calculation is made. Now we will come to the first term of solution chemistry which is known as normality. It is the term that is generally used to express the strength of a given solution. So what do you mean by normality? Its definition is that when a gram equivalent weight of solute is dissolved in one liter of solvent then the obtained solution is called as one normal solution. So the definition is very simple, when we prepare a solution by dissolving any solute but how much solute we are taking, its gram equivalent weight and how much volume we are taking, we are taking the volume as one liter. Then the resultant solution is called as one normal solution. So the same thing we are expressed with the help of equation normality of a solution is equal to gram equivalent weight of solute divided by volume of solution in liter. Therefore how to calculate gram equivalent weight of solute? It is nothing but weight of solute in gram divided by equivalent weight of that particular solute. So the modified state becomes normality of solution is equal to weight of solute into gram which is expressed as W s divided by equivalent weight of solute which is expressed as E s multiplied by volume of solution in liter always it is correlated to one liter. Therefore equivalent weight how it is calculated because we need to keep the value of E. So equivalent weight of any solute is obtained by dividing molecular weight of that solute with either valency number or the number of electrical charges. So valency number is applicable in case of salt. For example NaCl, Na is having one charge plus Na plus and Cl is having one charge that is minus that is valency number or electrical charges in case of salt and if the material is acid we are dividing the weight divided by its number of H plus ion. So finally normality n is equal to W upon E multiplied by V in liter where W express weight of solute, E express equivalent weight of that solute and V is nothing but volume of that solution in liter. Now I would like to ask you one question here. What is the equivalent weight of magnesium sulfate? I think for a moment we will come with answer. The correct answer for a given question is that equivalent weight of magnesium sulfate is 60. First of all we will calculate the molecular weight of magnesium sulfate. Molecular weight is the sum of atomic weights of the atom which are present in that molecule. So in a given molecule magnesium is one atom, sulfur is another atom and oxygen is one more atom. So what is the weight of magnesium 24? What is the weight of sulfur 32 and what is the one atom weight of oxygen that is 16 multiplied by 4? So its molecular weight becomes 120 and we know that its valence is 2 because when it dissociate it forms mg plus plus and SO 4 minus minus. So the number of electrical charges are 2. So I will divide 120 by 2 and the correct answer is 60. Now we will see the second term that is molarity. How it is defined? It is another concept when one gram mole of solute is dissolved in one liter of solvent then the obtained solution is called as one molar solution. So how it is prepared? It is prepared by dissolving a gram mole of solute in a one liter of solvent either it may be water or it may be non aqueous solution. Now what is its equation? Molarity of solution is equal to number of moles of solute divided by volume of solution in liter. So molarity of solution M is equal to N upon V in liter. So what is N? N is number of moles. Similarly one more term is there that is molality. The difference in between molarity and molality is that in case of molarity we are taking the solvent in one liter that is volume and in case of molality instead of taking as a volume we are taking as a mass for that solvent and it becomes one kg. That is the only difference. Now what is its definition? When a gram equivalent weight of solute is dissolved in one kg of solvent then the obtained solution is called as one molal solution. What is molality? Molality is a concept where a solution is prepared by dissolving a gram equivalent weight of solute in one kg of solvent. So molality of solution is equal to number of moles of solute divided by weight of solvent in kg. After learning the theoretical part now we will apply that and we will do the practice how to calculate the molality and molarity of a given solution. So this is the problem number one. Calculate normality of a solution containing 9.8 gram of sulfuric acid in one liter of water where the given data is that the molecular weight of sulfuric acid is 98. The given data is that solute sulfuric acid its molecular weight is 98. Combine of solution is one liter weight of solute is equal to 9.8 gram. So normality N is equal to W divided by E into V. So we will keep the value of W that is 9.8 equivalent weight here we have the compound sulfuric acid and therefore 98 is needed to divide by 2 its equivalent weight becomes 49. So the final answer becomes 0.2 normal. Therefore normality of a given solution is 0.2 normal. We will see one more problem. Problem number two. Calculate normality of a solution containing 18.2 gram of hydrochloric acid in one liter of water the same as we have learned in problem number one this is a given data hydrochloric acid 36.5 volume is 1 liter and weight of the solute is 18.2. So by putting the value we will get the answer 0.5 and therefore normality of a given solution is 0.5 normal. Now this is one more different kind of problem. Calculate the weight required to prepare 0.5 normal 500 ml koH solution. So here instead of asking to calculate either normality or molarity we have been asked to calculate what is the weight required to prepare a said normal a said volume solution and in this example the solute is koH, molecular weight is 56, volume of the solution is 500 ml we have to convert into liter which becomes 0.5 liter and normality of given solution is 0.5 normal. So it is a standard formula n is equal to w upon e multiplied by v and the modified formula becomes weight of solute is equal to n multiplied by e multiplied by v by putting the value will get the answer 14 gram therefore weight of solute to required to prepare a given solution is 14 gram. One more problem we will see calculate the weight required to prepare 0.05 molar 500 ml mgCl2 solution. Here the given solute is mgCl2 given molarity is 0.05 so by using this given data and by using the standard formula of molarity is equal to ws that is weight of solute divided by ms that is molecular weight of solute multiplied by v in liter. So what happens molarity is given and we have been asked to calculate ws so the modified formula becomes ws that is weight of solute is equal to m multiplied by ms multiplied by v so by putting the value we are getting the answer 2.35 gram therefore the weight of the solute required to prepare the said solution is 2.35 gram.