 Good morning, myself Dr. Nitin Gramopadhyay working as assistant professor in Department of Humanities and Sciences. Today we are going to learn topic analytical chemistry and a particular beat from that topic is solution chemistry. This is the learning outcome for today's session. At the end of this session, students will be able to define the terms of solution chemistry and apply the knowledge to solve the problems. This is the content we are going to learn in this session. First, we will learn the definitions of normality, molarity, molality and then we will learn how to solve the numericals to find out the normality, molarity and molality of a given example. What do you mean by normality? The term normality is defined as the number of gram equivalents of the solute dissolved in a suitable solvent and then the solution is diluted to 1 liter. Then normality of the resultant solution is one normal. What is one normal solution? When a solution is prepared by dissolving a number of gram equivalents of the solute in 1 liter of volume, so normality is equal to number of gram equivalents of the solute divided by volume of the solution in liter. Therefore, how to find out the gram equivalent weight of the solute that is weight of solute in grams that is denoted by W s divided by equivalent weight of solute E s multiplied by volume of solution in liter that is V in liter. So, normality n becomes W upon E multiplied by V in liter where W is weight of the solute in grams, E is equivalent weight of the solute and V is volume in liter. Now, we will see molarity. The term molarity is defined as the number of gram moles of the solute dissolved in a suitable solvent and then the solution is diluted to 1 liter. Then the resultant solution is known as one molar solution. So, what is the basic difference in between normal and molarity? The normal solution is prepared by dissolving the gram equivalent weight of the solute and molar solution is prepared by dissolving the gram moles of the solute in 1 liter of the volume. So, molarity m is equal to gram moles of the solute divided by volume of the solution in liter. Therefore, now molarity of solution m is equal to W s upon M s multiplied by 1 upon V in liter where W s is weight of solute, M s is molecular weight of solute and V is volume in liter. So, we require to know what is the gram mole of the solute is dissolved and gram mole of the solute is nothing but the weight of solute taken divided by molecular weight of that particular solute. So, this is about molarity. Now, what about molality? Molality is denoted by small m and there is only one difference in between molarity and molality. In case of molarity, the volume is measured in liter and in case of molality, the volume is measured in kg. So, the term molality is defined as the number of gram moles of the solute dissolved in a suitable solvent and then the solution is diluted to 1 kg, then molality of the solution is 1. So, we have learned in case of molarity, in case of normality the volume is 1 liter, but in case of molality it is 1 kg. So, we are dissolving the gram mole of the solute in 1 kg of the solvent. So, molality m is equal to gram moles of the solute divided by volume of solution in kg. The same thing is there that is molality is equal to W s upon m s multiplied by 1 upon W that is solvent in kg. So, this W s is weight of solute, m s is molecular weight of solute and this W is weight of the solvent and that is correlated to 1 kg. Now, here I would like to ask you one question. The gram moles of the solute dissolved in 1 kg of solvent, then the obtained solution is called as what is the question? The gram moles of the solute dissolved in 1 kg of solvent, then the obtained solution is called as option A 1 normal solution, option B 1 molar solution, option C 1 molar solution and option D all of these. Solving for a moment and the correct answer is option C that is 1 molal solution. When 1 gram mole of solute is dissolved in 1 kg of solvent, then the obtained solution is called as molal solution because the concept of molarity and molality gets varied only on the volume or the mass. In case of molar the solvent is measured in volume that is 1 liter and in case of molal the solvent is taken in mass that is 1 kg. So, the correct answer to this question is option C that is 1 molal solution. Now, after learning the definition of normality, molarity and molality, we will apply the same knowledge to solve the numerical based on to calculate either normality or molarity or what is the weight we require to prepare the solution. So, the first question is that what is the weight of M G SO 4 that is magnesium sulphate required to prepare 0.5 normal 500 ml aqueous solution. So, the given conditions is that the solute name is magnesium sulphate. We have to prepare a solution of 0.5 normal and we have to prepare that solution the volume given is that 500 ml. So, as this is related with the normality we will write first the normality formula and the normality is equal to W upon E multiplied by V in liter where N is normality of a given solution that is given for this example that is 0.5 normal. V is equal to volume of a given solution that is 500 ml, but as per our definition we require the volume in liter so it becomes 0.5 liter and E equivalent weight of solute given. So, how to calculate the equivalent weight E is equal to equivalent weight of a given solute that is is equal to molecular weight of the solute divided by valances of magnesium or the sulphate ion so that is 2 answer is that. So, we know the molecular weight of the magnesium sulphate is 120 that is divided by 2 so it becomes 60 so 60 is a equivalent weight for a given solute. Now, the same thing I will keep here the formula is that what is the weight we require so the modified formula becomes W is equal to E multiplied by N multiplied by V in liter. So, what is E equivalent weight of a given solute that is 60 N normality for a given example that is 0.5 and V that is volume of a solution in liter that is 0.5 and the answer is 15 gram. Therefore, for a given example 15 gram of magnesium sulphate is required to prepare 0.5 normal 500 ml aqueous solution. So, this is the way how to calculate what is the weight we require to prepare a particular solution of a particular volume. Now, we will learn again one more problem in first problem we have learnt by knowing the normality and volume how to calculate the weight required. But in second numerical we will learn what is the normality of a solution if we know what is the weight of solute we have dissolved and the volume we know. So, the question is that what is the normality of a 500 ml solution prepared by dissolving 7.2 gram of HCl in water. So, the solution is that the solution is 500 ml in volume it becomes 0.5 liter, but that is obtained by dissolving 7.2 gram of HCl in water and we have been asked to calculate what is its normality. So, I will write the formula normality is equal to N is equal to W upon E multiplied by V in liter where N is normality of a given solution that is needed to calculate V volume of a given solution that is 500 ml is equal to 0.5 liter and E equivalent weight of a given solute. Now, again by the same method we will calculate what is the equivalent weight for this given solute. The formula we know that is molecular weight of solute divided by valances of H plus ion or the chlorine ion and the valance is only 1. Therefore, the molecular weight of the HCl is 36.5 which is divided by 1 is equal to 36.5 and the weight is given that is 7.2 gram. What is the normality formula? N is equal to W upon E s multiplied by V in liter. So, W is weight of solute given that is 7.2 divided by E s is 36.5 multiplied by V 0.5 liter and the answer is 0.394 normal. So, therefore, normality of a given solution is 0.394. So, in the second example we have learned by knowing the weight of solute by knowing the equivalent weight of solute by knowing the volume of a solution how to calculate the normality and the answer for the given example is 0.394. All this I have prepared by using a textbook of engineering chemistry written by Jain and Jain. Thank you.