 Our today's topic of discussion is analytical chemistry, numericals. So at the end of this session, students will be able to solve the numericals based on normality, molarity and molality. Content of our today's lecture, we are going to learn about normality, its definition and numerical, molarity, its definition and numerical, and molality, its definition and numerical. So normality. Normality is represented by N. The number of gram equivalents of the solute per liter of solution is the normality of solution. Mathematically it can be represented as normality of solution is equal to gram equivalent of solute divided by volume of solution in liters. So normality of solution we can calculate with the help of formula that is weight of solute in grams that is W s divided by equivalent weight of solute that is E s multiplied by volume of solution in liters that is V liters. So normality in general is equal to W upon E into V in liters, molarity. Molarity is represented by capital M. It is defined as the number of moles of the solute per liter of solution is the molarity of solution. Mathematically it can be represented as molarity of solution is equal to number of moles of solute divided by volume of solution in liters that is M is equal to N by V in liters where N is number of moles we can calculate the number of moles of solute by using the formula that is weight of solute divided by molecular weight of the solute. So overall capital M that is molarity is equal to weight of solute divided by molecular weight of solute into 1 by V in liters. Molality is represented by small m. It is defined as the number of moles of the solute per kilogram of solvent is the molarity of solution. So molarity of solution mathematically we can represent that that is small m is equal to number of moles of solute divided by the weight of solvent in kilograms. So N stands for the number of moles overall that is why small m that is molarity is equal to weight upon molecular weight into weight of solvent in kilograms. Let us solve some numericals based on this normality molarity and molality. So first based on normality 5.6 grams of calcium carbonate is dissolved in small volume of water using dilute hydrochloric acid and then dilute it to final volume of 500 ml. Calculate the normality of the solution. Here in this case the formula to be used is normality is equal to weight upon equivalent weight into volume in liter where weight it is given 5.6 grams. Volume is 500 ml that is 0.5 liters and equivalent weight of the calcium carbonate we can calculate by using the formula that is molecular weight upon charge on either calcium or carbonate. Therefore molecular weight of this calcium carbonate is 100 divided by 2 because the charge on calcium or carbonate is 2 and therefore answer is 50 equivalent weight of calcium carbonate is 50. So by putting all these values in the above equation we can calculate the normality that is n is equal to 5.6 divided by 50 into 0.5. That answer is normality is equal to 0.22 normal. Next numerical 4.4 grams of sodium hydroxide is dissolved in water to make 1200 ml of the solution. Calculate the normality of the solution. The formula to be used once again that is normality is equal to weight upon equivalent weight into volume in liter. Given 4.4 grams of the solute is given that is W is equal to 4.4 grams. Volume is given that is 1200 ml that is 1.2 liters and equivalent weight of the sodium hydroxide we can calculate that is 40 as in case of sodium hydroxide the molecular weight and equivalent weight are same. So by putting all these values in the equation of normality calculation normality is equal to 4.4 divided by 40 into 1.2 that comes as a 0.091 normal. So the normality of the given solution is 0.091 normal. Next numerical what is the weight of calcium chloride required to prepare 700 ml of 0.05 normal solution. Molecular weight of the calcium chloride it is equal to 111 that is 111. So here in this case the modified formula to be used is weight of solute required that is W is equal to normality into equivalent weight into volume in liter. Hence the given data is likewise normality is given 0.05 normal. Volume of solution it is 700 ml that is 0.7 liters and equivalent weight of the calcium chloride is 111 divided by 2 that comes as a 55.5. By putting all these values in the equation that is W is equal to 0.05 into 55.5 into 0.7 that comes as W that is weight required is 1.94 grams second molarity. So calculate the molarity of the solution containing 0.5 grams of sodium hydroxide dissolved in 500 ml of the solution. So here in this case the formula to be used is molarity capital M is equal to weight upon molecular weight into volume in liters. So given data weight of solute is given that is 0.5 grams. Molecular weight of solute is 40 and volume is 500 ml that is 0.5 liters. So by putting all these values in the given equation that is capital M is equal to weight 0.5 divided by molecular weight that is 40 into volume that is 0.5. So by calculation we are getting the answer that is molarity of the sodium hydroxide solution it is 0.025 molar. Just numerical calculate the weight of oxalic acid molecular weight 126 required to prepare 700 ml of 0.2 molar solution. So formula to be used that is weight of solute required is equal to molarity of solution into molecular weight of solute into volume in liters. So the molarity is given that is 0.2 molecular weight of the solute is 126 and V that is 700 ml that is 0.7 liters. So by putting all these values in this particular equation we can get that is W is equal to 0.2 into 126 into 0.7. So the answer is 17.64 grams. So pause the video now and answer this question. One is how much silver nitrate molecular weight to 170 required to prepare 500 ml of 0.2 normal and 0.2 molar solution options are A 8.25 and 4.25 grams B 4.25 and 8.25 grams C 17 and 17 grams and D none of this welcome back. So the answer for the normality and molarity solution is C 17 and 17 grams molality. Calculate the molality of solution containing 19.6 grams of sulfuric acid in 1 kilogram of solvent. So here in this case the formula to be used is small m that is molality is equal to weight upon molecular weight into weight of solvent in kilograms given data weight is given that is 19.6 grams molecular weight is 98 and weight of the solvent is 1 kilograms. So by putting all these values in this equation we can calculate m is equal to 19.6 upon 98 into 1 it is equal to 0.2 molar. So the answer is molality is equal to 0.2 molar references a textbook of engineering chemistry by Jane and Jane I have used for this particular topic thank you.