Units of Measurement PDF

10 - UNITS OF MEASUREMENTS Fundamental Quantities: Physical quantities which are fundamental in nature, which are independent of other quantities are called as fundamental quantities. Ex: Length, mass, time, thermdynamic temparature etc. Units: Quantity used as standard of measurement is...

10 - UNITS OF MEASUREMENTS Fundamental Quantities: Physical quantities which are fundamental in nature, which are independent of other quantities are called as fundamental quantities. Ex: Length, mass, time, thermdynamic temparature etc. Units: Quantity used as standard of measurement is called as unit. Fundamental Units : Units used to measure the fundamental quantities are called fundamental units. Ex: Centimetre for length in C.G.S. system Gram for mass in C.G.S. system Second for time in in C.G.S. system Derived quantities: Quantities which are derived from fundamental quantities are called derived quantities. Ex: Area, Volume, density, work. Derived units : Units used to measure derived quantities are called derived units. Ex: Cm2 in C.G.S. system for area Cm3 in C.G.S. system for volume gm Cm-3 in C.G.S. system for density. Systems of Units : There are several systems of units for measurement of fundamental quan- tities. 1. British system or F.P.S. system 2. Metric system or C.G.S. system 3. M.K.S. system 141 1. F.P.S. system In British system, units of length, mass and time are Foot 5. Acceleration metre / second2 ms-2 (f), pound (p) and second (s) respectively. 6. Momentum Kilogram metre/second Kg m s-1 2. C.G.S. system : In metric system, units of length mass and time are 7. Force Newton N centimetre (m), gram (gm), and second (s) 8. Impulse Newton second NS 3. M.K.S. system : In this system, units of length mass and time are metre(m), kilograms (kg) and second (s). In M.K.S. system current 9. Work & Energy Joule J is taken as fourth fundamental quantity and its unit is ampere. 10. Power Watt W Hence this system is also called as M.K.S.A. system. 11. Pressure Pascal Pa S.I. Units : It is also called as international system of units. 12. Coefficient of viscosity Poiseuille Pi There are seven fundamental and two supplementary quantities in 13. Frequency Hertz Hz international system of units. 14. Surface tension Newton / metre Nm-1 Fundamental Quantities and their Units in S.I. Units : 15. Specific heat J / Kg / k J Kg-1 k-1 Fundamental Quantity Unit Symbol 16. Latent heat J / Kg J Kg-1 1. Length Metre m 17. Wave length metre m 2. Mass Kilogram Kg 18. Wave number 1/metre m-1 3. Time Second S Definitions of fundamental, supplementary and some derived units of S.I.: 4. Thermo dynamic temparature Kelvin K Fundamental units of SI and their definitions : 5. Lumincus intensity Candila Cd 1. Metre (m) : Metre is the distance travelled by light in vaccum in 1 / 6. Electric current Ampere A 299,792,458th of a second. It is the S.I. unit of length. 7. Quantity of substance Mole mol 2. Kilogram (Kg) : Kilogram is the mass of a cylinder of platinum - Iridium Supplementary quantities and their units in S.I. alloy kept at international bureau of weights and measures at paris. It 1. Plane angle Radian Rad is S.I. unit of mass. 2. Solid angle steradian sr 3. Second (S) : Second is the time interval taken by 9, 192, 631, 770 cycles of certain radiation emitted by caesium133 atom in ground Some of the derived quantities and their units in S.I.: state. It is the S.I. unit of time. 1. Area (meter)2 m2 4. Kelvin (K) : Kelvin is 1 / 273.16 th of thermodynamic temparature 2. Volume (meter)3 m3 of triple point of water. Kelvin is the S.I. unit of temparature. 3. Density kilogram/meter3 Kg m-3 5. Candila (Cd) : Candila is the luminous intensity in a direction 4. Velocity metre / second ms -1 normal to surface area of 1 / 600000 m2 of a black body at the 142 143  11. kilo 103 k freezing point of platinum at a pressure of 101325 Nm-2. It is the S.I. 6 unit of luminous intensity. 12. Mega 10 M 9 6. Ampere (A) : Ampere is the current which flowing through in each of 13. Giga 10 G two straight parallal conductors of infinite length and negligible cross 12 14. Tera 10 T section placed at a distance of 1 metre apart in vaccum produces a 15 force of 2 x 107 newton per metre on each conductor. 15. Peta 10 P 18 7. Mole (mol) : Mole is the amount of substance which contains as many 16. Exa 10 E elementary entities (molecules or atom) as there are atoms in 0.0012 Unit of volume in S.I : Kg of C12. S.I. Unit of volume is 1metre3. Supplementary units and their definitions. Accepted unit of volume is 1 litre. 1. Radian (rad) : Radian is the angle subtended at the centre of a circle 1 Litre : 1 Litre is the volume occupied by one kilo gram of water of radius 1 metre by an arc of length 1 metre. It is the S.I. unit of at 40 C plane angle. 1ml : 1 ml is the volume occupied by 1 g of water at 40 C. 2. Steradian (sr) : Steradian is the angle subtended at the centre of a sphere of radius 1 metre by the surface area of 1m2. It is the S.I. Some useful conversions of mass,length and volume : unit of solid angle. Mass : Prefixes in S.I. units : 1 Kg =1000 gm 1 gm =10-3 kg. S.No. Prefix Power of 10 Symbol 1 gm =1000 mg 1 mg =10-3 g. -1 1. deci 10 d 1 mg = 1000 mg 1 mg = 10-3 mg. 2. centi 10-2 c 1 mg = 1000 ng 1 ng = 10-3 mg. 3. milli 10-3 m 1 kg = 103 g. = 106 mg = 109 ng 4. micro 10-6 m Length : 5. nano 10-9 n 1 metre = 102cm 1cm = 10-2 metre. -12 6. Pico 10 p 1 metre = 103mm. 1mm = 10-3 metre. -15 7. Fento 10 f 1 metre = 10-6 mm 1 mm = 10-6 metre. -18 8. Atto 10 a 1 metre = 1010A0 1A0 = 10-10 metre = 10-8 cm. 1 9. deca 10 da 1 metre = 1011 X-ray units 1X ray unit =10-11 metre. 10. hecto 102 h 1 metre = 1015 Fermi 1 Fermi = 10-15 m 144 145  1. What are fundamental quantities? Give examples. 1 metre = 102 cm = 103 mm = 106 mm = 109 nm 2. What are units? 1010A0 = 1011 X-ray units = 1015 Fermi 3. Define fundamental uints and give examples. 4. Define derived quantities and exemplify. Volume : 5. What are derived units and exemplify? 1 Litre = 1000 ml. 1 ml = 10-3 Litre 6. What is F.P.S. (British) system? 1ml = 1000 mL 1 mL = 10-3 ml. 7. Mention the fundamental quantities and their units in C.G.S. (Metric) system. 1 deci litre = 100 ml. 8. Write the fundamental quantities and their units in M.K.S. system. Measures of Volume in Imperial System 1 Gallon (C) = 8 pints 9. Mention the fundamental quantities in international system (S.I.) 1 pint (O) = 20 fluid ounces 10. What are the units of fundamental quantities in S.I.? 1 fluid ounce ( 37 ) = 8 fluid drachms 11. What are supplementary quantities and their units in international system (S.I.)? 1 fluid drachm ( 3 ) = 60 minims (m) 12. Mention any two derived quantities and their units in nternatioal system Conversion with Metric System (S.I.). 1 fluid ounce ( 37 ) = 30 ml. 13. Define “meter”. 1 pint (O) = 20 fluid ounces = 20 X 30 = 600 ml. 14. Define “Kilogram”. 1 Gallon (C) = 8 pints = 8 X 20 X 30 = 4,800 ml. 15. Define “second”. 16. Mention the unit of volume in S.I. and define. SUMMARY Fundamantal quantities are physical quantities which are independent of other quantitites. Unit is a standard quantity used for measurement. Derived quantities are derived from fundamental quantities. Different systems of units are British system, Metric system, MKS system and S.I. Essay Question 1. Write about different systems of units. Short Answer Questions 146 147  11. SOLUTIONS 8. Liquid in gas solution : Liquid solute dissolved in gaseous solvent is called liquid in gas solution. Solution : It is defined as homogeneous mixture of two or more substances. Ex: Moisture in air Ex: 0.9% Sodium Chloride solution 9. Gas in gas solution : Gaseous solute dissolved in gaseous solvent is Solute : Substance present in minor proportion is called as solute. called gas in gas solution. Ex: Sodium chloride present in 0.9% Sodium Chloride solution Ex: Air Solvent : Substance present in major quantity is called as solvent. Ex: Water present in 0.9% Sodium Chloride solution Methods of expressing Concentration of Solutions : Different types of solutions : Different methods of expressing concentration are Based upon the states of matter of solute and solvent, solutions can be classified 1. Percentage 2. Molarity 3. Normality into 4. Molality 5. Formality 6. Molefraction 1. Solid in liquid solution : Solid solute dissolved in liquid solvent is called solid in liquid solution. Types of solutions based on the mode of expression of concentration : Ex:: 10% sodium tungstate solution 1. Percentage solutions : Solutions whose concentration are expressed as percentage are called as percentage solutions. 2. Liquid in liquid solution : Liquid solute dissolved in liquid solvent is called liquid in liquid solution. 2. Molar solutions : Solutions whose concentrations are expressed in terms of molarity are molar solutions. Ex: 2% Acetic acid solution 3. Gas in liquid solution : Gaseous solute dissolved in liquid solvent is called 3. Normal solutions : Solutions whose concentrations are expressed in gas in liquid solution. normality are normal solutions. Ex: Aerated water. 4. Molal solutions : Solutions whose concentrations are expressed in molality are molal solutions. 4. Solid in solid solution : Solid solute dissolved in solid solvent is called solid in solid solutions. Percentage Solutions : Ex: Alloys Solutions whose concentrations are expressed as percentage are called as 5. Liquid in solid solution : Liquid solute dissolved in solid solvent is called percentage solutions. liquid in solid solution. There are four kinds of percentage solutions. Ex: Hydrated salts such as Na2CO3,10 H2O 1. Weight in volume percentage solutions (w/v solutions) 6. Gas in solid solution : Gaseous solute dissolved in solid solvent is called 2. Weight in weight percentage solutions (w/w solutions) gas in solid solution. 3. Volume in weight percentage solutions (v/w solutions) Ex: H2 in palladium 4. Volume in volume percentage solutions (v/v solutions) 7. Solid in gas solution : Solid solute dissolved in gaseous solvent is called solid in gas solution. Generally w/v solutions and w/w solutions are percentage solutions of Ex: Minute particles in inhaled air solids in liquids. 148 149  v/w solutions are percentage solutions of liquids in solids. 100 ml. ------- 10 grams v/v solutions are percentage solutions of liquids in liquids. 250 ml. ------- ? 10 X 250 1. Weight in volume percentage solutions (w/v solutions). = 25 grams. w/v solutions are solutions of solids in liquids, one part of solid solute by 100 weight dissolved in liquid solvent to produce solution of 100 parts by 25 grams of sodium tungstate dissolved in enough water and diluted to 250 volume is called as 1% weight in volume solution. ml. with water will produce 10% w/v sodium tungstate solution. General Formula : Solute - 1 g. Method of Preparation : Solvent up to 100 ml. } will give 1% w/ 1. Weigh accurately 25 grams of sodium tungstate. v solution. 2. Dissolve in three quarters of the volume of water (approximately 168ml.). Ex: Prepare 200 ml. of 0.9% w/v sodium chloride solution. 3. Examine for the presence of foreign particles. If foreign partcles are visible, filter through a plug of cotton wool / filter paper placed in a funnel. 100 ml. ------- 0.9 G 4. Dilute to 250 ml. with distilled water in a graudated measuring jar. 200 ml. ------- ? 0.9 X 200 5. Transfer into a clean and dry container. = 1.8 G. 6. Label neatly. 100 Weight in Weight solutions : 1.8 Grams of sodium chloride dissolved in enough water and diluted to 200 ml. with water will produce 200 ml. of 0.9% w/v sodium chloride W/w solutions are also % solutions of solids in liquids. One part of solid solution (Normal saline). solute by weight dissolved in liquid solvent to produce solution of 100 parts by weight in called as 1% weight in weight solution. Method of Preparation : General formula : Solute - 1 g. 1. Weigh accurately 1.8 grams of sodium chloride. } will give 1% w/w solution. Solvent up to 100 g. 2. Dissolve in about three quarters of the solvent (slightly less than 150 ml. of water). w/v solutions are more common in application among % solutions of solids in liquids. 3. Examine for presence of foreign particles by holding against light. If foreign particles are visible, filter with a plug of cotton wool / filter paper Volume in weight solutions (v/w solutions) : v/w solutions are percentage placed in a funnel. solutions of liquids in solids. One part by volume of liquid solute dissolved in solid to produce solution of 100 parts by weight is called as 1% volume 4. Dilute to 200 ml. volume with water in a graduated measuring jar. in weight solution. 5. Transfer into a clean and dry container. General formula : Liquid - 1 ml. 6. Label neatly. Solid up to 100 g. } will give 1% v/w solution. Ex: 2. : Prepare 250 ml. of 10% w/v sodium tungstate solution. 150 151 Volume in volume solutions (v/v solutions) : Method of Preparation : v/v solutions are percentage solutions of liquids in liquids. One part by 1. Take 15 ml. of formalin in a measuring jar. volume of liquid solute diluted to 100 parts by volume with liquid solvent 2. Dilute to 150 ml. with water. is called as 1% volume in volume solution. General formula : Liquid - 1 ml. 3. Transfer into a clean and dry container. Solvent up to 100 ml. } will give 1% v/v solution. 4. Label neatly. Ex:1 Prepare 50 ml. of 2% v/v acetic acid Molar solutions : Molar solution is defined as a solution containing one mole of substance in one litre of solution. 100 ------- 2 Molarity : Molarity is number of moles of the solute per litre of solution. It is 50 ---------- ? denoted by M. 50 X 2 = 1 ml. number of moles of solute M= 100 volume of solution in litres 1 ml. of acetic acid diluted to 50 ml. with water will produce 50 ml. of 2% Mole : Mole is defined as one gram molecular weight substance present in one v/v acetic acid solution. litre solution. Method of preparation: Gram molecular weight : Molecular weight expressed in grams is called as 1. Take apporoximately 45 ml. of water into a measuring jar. gram molecular weight. 2. Add 1 ml. of acetic acid with stirring. Ex: One mole of H2SO4 = 98.078 grams of H2SO4 in one litre solution. 3. Make the volume to 50 ml. with water. One mole of HCl = 36.5 grams of HCl in one litre solution. 4. Transfer to a clean and dry container. Method of Preparation of 1 M solution : 5. Label neatly. 1. Weigh accurately quantity equivalent to one gram molecular weight of the Ex: 2 Prepare 150 ml. of 10% (v/v) formalin in normal saline. substance. 100 ------- 10 2. Dissolve in approximately 750 ml. of water. 150 --------- ? 3. Examine for foreign particles and filter if necessary. 150 X 10 4. Dilute to 1 litre with water in a graduated measuring jar. = 15 ml. 100 5. Transfer into a clean and dry container. 15 ml. of formalin diluted to 150 ml. with water will produce 150 ml. of 6. Label neatly. 10% v/v formalin. 152 153  Substance Molecular weight mass of one mole in one litre Method of preparation: ie. Gram molecular weight 1) Take accurately quantity of HCl equal to 0.273 g of HCl. H2SO4 98.078 98.078 grams 2) Gradually add to about 50 ml. of water with stirring and cool. HCl 36.5 36.5 grams 3) Dilute to 75 ml. with distilled water. 4) Standardise and adjust molarity if necessary. Molarities of commercially available concentrated acids : Problem-2 : Commercially available concentrated hydro-chloric acid has 10.5- 12 M concentration. Prepare 100 ml. of 0.5 M Hcl. with commerciallly available HCl. Concentrated sulfuric acid is about 18 M. (specific gravity of commercially available HCl = 1.16) These two acids are widely used in the preparation of standard solutions of percentage by weight = 36% w/w acids. Solution : Molecular weight of HCl = 36.5 Problems : Gram molecular weight of HCl = 36.5 g. Problem-I : i.e., weight of HCl required to make 1 litre of 1 M HCl solution = Prepare 75 ml. of 0.1 M Hcl. 36.5 g. Solution : Molecular weight of HCl = 36.5 Weight of HCl required to make 1 litre of 0.5M HCl = Gram molecular weight of HCl = 36.5 g. 1 M -------- 36.5 g i.e. weight of required to make 1 ltr. of 1 M HCl = 36.5 g. weight of HCl required to make 1 litre 0.1 M HCl = 0.5 M --------- ? 0.5 X 36.5 1 M ------- 36.5 g = 18.25 g. 0.1 M --------- ? 1 0.1 X 36.5 Weight of HCl required to make 100 ml. of 0.5 M HCl = = 3.65 g. 1 1000 ml. -------- 18.25 g weight of HCl required to make 75 ml. of 0.1 M solution = 100 ml. --------- ? 1,000 ml. -------- 3.65 g 100 X 18.25 = 1.825 g. 75 ml. --------- ? 1000 3.65 X 75 Weight of HCl required to make 100 ml. of 0.5 M HCl = 1.825 g. = 0.273 g. 1000 Quantity of commercially available concentrated HCl required to make... weight of HCl required to make 75 ml. of 0.1 M HCl = 0.273 g. 154 155  i.e. Weight of H2SO4 required to make 1 litre of 0.1 M H2SO4 = 100 ml. of 0.5 M HCl = 1 M -------- 98 g percentage by weight of HCl = 36% w/w, 0.1 M --------- ? 36 g. -------- 100 g 0.1 X 98 1.825 ---------- ? = 9.8 g. 1.825 X 100 1 = 5.07 g. Weight of H2SO4 required to make 500 ml. of 0.1 M H2SO4 = 4.9 g. 36 Quantity of commercial concentrated H2SO4 required to make 500 ml. of specific gravity of HCl = 1.16 0.1 M H2SO4 = 1.16 g. -------- 1 ml. percentabe by weight of H2SO4 = 95% w/w 5.07 ---------- ? 95 g. -------- 100 g 5.07 X 1 = 4.37 ml. 4.9 g --------- ? 1.16 4.9 X 100. = 5.16 g... Quantity of commercial concentrated HCl required to prepare 100 95 ml. of 0.5 M Hcl = 4.37ml. Weight of commercial concentrated H2SO4 required to make 500 ml. of Method of Preparation : 0.1 M H2SO4=5.16g. 1) Measure accurately 4.37 ml. of commercial concentrated HCl. specific gravity of H2SO4 = 1.84 2) Gradually add to about 75 ml. of distilled water with stirring and 1.84 g -------- 1 ml. cool. 5.16 g --------- ? 3) Dilute to 100 ml. with distilled water. 5.16 X 1 4) Standardize and adjust molarity if necessary. = 2.8 ml. Problem - 3 : 1.84 Prepare 500 ml. of 0.1M H2SO4 solution with commercially available... Quantity of commercial concentrated H2SO4 required to make 500 ml. concentrated H2SO4 (specific gravity of commercial of 0.1 M H2SO4 =2.8ml. concentrated H2SO4 = 1.84, Method of Preparation : percentage by weight = 95% w/w) 1. Measure accurately 2.8 ml. of commercial concentrated H2SO4. Solution : Molecular weight of H2SO4= 98 2. Gradually add to about 450 ml. of distilled water with stirring and cool. gram molecular weight of H2SO4= 98 g. 3. Dilute to 500 ml. with distilled water. Weight of H2SO4 required to make 1 litre of 1 M H2SO4 = 98 g. 4. Standardise and adjust molarity if necessary. 156 157 Problem - 4 : Quantity of commercial concentrated HCl required to make 500 ml. of 0.2 Prepare 50 ml. of 1M NaOH M HCl. = 60 ml. -------- 1 ml. Solution : 500 ml. --------- ? Molecular weight of NaOH = 40 500 X 1 = 8.3 ml. Gram molecular weight of NaOH = 40 g. 60 i.e. Weight of NaOH required to make 1 litre of 1 M NaOH solution=40g. Quantity of commercial concentrated HCl required to make 500 ml. of 0.2 M HCl = 8.3ml. Weight of NaOH required to make 50 ml. of 1 M NaOH = Method of Preparation : 1000 ml. -------- 40 g. 1. Measure accurately 8.3 ml.of commercial concentrated HCl. 50 ml. --------- ? 50 X 40 2. Gradually add to about 400 ml. of distilled water, mix and cool. = 2 g. 3. Dilute to 500 ml. with distilled water. 1000 4. Standardise and adjust molarity if necessary. Weight of NaoH required to make 50 ml. of 1 M NaOH = 2 g. Normal Solutions : Normal solution is defined as a solution containing one Method of Preparation : equivalent of substance in one litre of solution. 1. Weigh accurately 2 g. of NaOH. Normality : Normality is defined as number of equivalents of substance per 2. Dissolve in about 40 ml. of distilled water. litre of solution. It is denoted by N. 3. Dilute to 50 ml. with distilled water. Number of equivalents of substance N = 4. Standardise and adjust molarity if necessary. Number of litres of solution Problem - 5 : Prepare 500 ml. of 0.2M HCl. Number of milli equivalents of substance (Molarity of commercially available concentrated HCl = 12 M) = Solution : Number of mililitres of solution Concentration of commercially available concentrated HCl = 12M Substance Molecular Gram Molecular Equivalent Concentration to be prepared by dilution = 0.2 M. weight weight weight Higher concentration 12 M HCl 36.5 36.5 g 36.5 g. Dilution factor = = = 60 Lower concentration 0.2 M H2SO4 98.078 98.078 g 49.039 g. i.e., 1 ml. of 12 M HCl diluted to 60 ml. gives 0.2 M HCl. NaOH 40 40 g 40 g. 158 159  Normalities of commercially available concentrated acids : 3) Dilute to 500 ml. with distilled water. Concentrated sulfuric acid is available in approximately 36 N strength. 4) Standardise and adjust normality if necessary. Concentrated Hydrochloric acid is available in approximately 12 N strength. Problem - 2 Glacial acetic acid is available in approximately 45 N strength. Prepare 200 ml. of 0.5 N H2SO4 Solution : Molecular weight of H2SO4 = 98 Problem -1 Gram molecular weight of H2SO4 = 49 g. Prepare 500 ml. of 0.5 N HCl. Equivalent weight of H2SO4 = 49 g. Solution : Molecular weight of HCl = 36.5 Weight of H2SO4 required to make 1 litre of 1 N H2SO4 = 49 g. Gram molecular weight = 36.5 g. Weight of H2SO4 required to make 1 litre of 0.5 N H2SO4 = Equivalent weight of HCl = 36.5 g. 1 N -------- 49 g. i.e., Weight of HCl required to make 1 litre of 1N HCl = 36.5 g. 0.5 N --------- ? Weight of HCl required to make 1 litre of 0.5 N HCl = 49 X 0.5 1 N -------- 36.5 g. = 24.5 g. 0.5 N --------- ? 1 0.5 X 36.5 Weight of H2SO4 required to make 200 ml. of 0.5 N H2SO4 = = 18.25 g. 1 1000 ml. -------- 24.5 g. Weight of HCl required to make 500 ml. of 0.5 N HCl = 200 ml. --------- ? 24.5 X 200 1000 ml. -------- 18.25 g. = 4.9 g. 500 ml. --------- ? 1000 500 X 18.25 Weight of H2SO4 required to make 200 ml. of 0.5 N H2SO4 = 4.9 g. = 9.125 g. 1000 Method of preparation :... weight of HCl required to make 500 ml. of 0.5 N HCl = 9.125g. 1) Weigh accurately quantity of H2SO4 equal to 4.9 g. of H2SO4. Method of preparation : 2) Add gradually to approxymately 175 ml. of distilled water with constant stirring and cool. 1) Take accurately quantity of HCl equivalent to 9.125 g of HCl. 2) Add gradually to approiximately 450 ml. of distilled water with constant 3) Dilute to 200 ml. with distilled water. stirring and cool. 4) Standardise and adjust normality if necessary. 160 161 Problem - 3 = 36% w/w) Prepare 250 ml. of 0.5 N NaOH Solution : Molecular weight of HCl = 36.5 Solution :Molecular weight of NaOH = 40 Gram molecular weight of HCl. = 36.5 g. Gram molecular weight of NaOH = 40 g. Equivalent weight of HCl = 36.5 g. Equivalent weight of NaOH = 40 g. Weight of HCl required to make 1 litre of 1N HCl = 36.5 g. Weight of NaOH required to make 1 litre of 1 N NaOH = 40 g. Weight of HCl required to make 1 litre of 0.5 N HCl = 1N -------- 36.5 g. Weight of NaOH required to make 1 litre of 0.5 N NaOH = 0.5 N --------- ? 1 N -------- 40 g. 0.5 X 36.5 0.5 N --------- ? = 18.25 g. 0.5 X 40 1 = 20 g. Weight of HCl required to make 250 ml of 0.5 N HCl. = 1 1000 ml. -------- 18.25 g. Weight of NaOH required to make 250 ml. of 0.5 N NaOH = 250 ml. --------- ? 1000 ml. -------- 20 g. 250 X 18.25 250 ml. --------- ? = 4.56 g. 250 X 20 1000 = 5 g. Weight of HCl required to make 250 ml. of 0.5 N HCl = 4.56 g. 1000 Weight of commercial concentrated HCl required to make 250 ml. Weight of NaOH required to make 250 ml. of 0.5 N NaOH = 5 g. of 0.5 N HCl = percentage by weight = 36% w/w Method of Prepartion : 36 g. -------- 100 g. 1. Weigh accurately 5 g. of NaOH. 4.56 g --------- ? 2. Dissolve in about 200 ml. of distilled water. 4.56 X 100 g. 3. Dilute to 250 ml. with distilled water. = 12.66 g. 36 4. Standardise and adjust normality if necessary. Weight of commercial concentrated HCl required to make 25 ml. of Problem - 4 : 0.5 N HCl = 12.66 g. Prepare 250 ml. of 0.5 N HCl Quantity of commercial concentrated HCl required to make 250 ml. of 0.5 N HCl. = (specific gravity of commercial concentrated HCl = 1.16, percentage by weight 162 163  Specific gravity of HCl = 1.16 1000 ml. -------- 24.5 g. 1.16 g -------- 1 ml. 500 ml. --------- ? 12.6 g --------- ? 500 X 24.5 12.6 X 1 = 12.25 g. = 10.9 ml. 1000 1.16 Quantity of commercial concentrated H2SO4 required to make 500 ml. of Method of Preparation : 0.5 N H2SO4 = 1. Measure accurately 10.9 ml. of commercial concentrated HCl. percentage by weight = 95% w/w 2. Gradually add to about 190 ml. of distilled water, mix and cool. Weight of commercial concentrated H2SO4 required to make 500 ml. of 3. Dilute to 250 ml. with distilled water. 0.5 N H2SO4 = 4. Standardise and adjust normality if necessary. 95 g. -------- 100 g. Problem 5 12.25 --------- ? Prepare 500 ml of 0.5 N H2SO4 12.25 X 100. (specific gravity of commercial concentrated H2SO4 = 1.84 = 12.89 g. percentage by weight = 95% w/w ) 95 Solution : Weight of commercial concentrated H2SO4 required to make 500 ml. of 0.5 N H2SO4 = 12.89 g. Molecular weight of H2SO4 = 98 Quantity of commercial concentrated H2SO4 required to make 50 ml. of Gram molecular weight = 98 g. 0.5 N HCl = Equivalent weight = 49 g. Specific gravity of H2SO4 = 1.84 Weight of H2SO4 required to make 1 litre of 1 NH2SO4 = 49 g. 1.84 g. -------- 1 ml. Weight of H2SO4 required to make 1 litre of 0.5 N H2SO4 = 12.89 g. -------- ? 1 N -------- 49 g. 12.89 X 1 0.5 N --------- ? = 7 ml. 49 X 0.5 1.84 = 24.5 g. Quantity of commercial concentrated H2SO4 required to make 500 ml. of 1 0.5 N H2SO4 = 7ml. Weight of H2SO4 required to make 500 ml. of 0.5 N H2SO4 = Method of preparation : 164 165  1. Measure accurately 7 ml. of commercial concentrated H2SO4. Relation between Molarity and Normality : 2. Gradually add to about 375 ml. of water, mix and cool. Molecular weight Normality = Molarity X 3. Dilute to 500 ml. volume with water. Equivalent weight 4. Standardise and adjust normality if necessary. Molal Solutions : Molal solution is defined as solution containing one mole Problem 6 of solute in one kilogram of solvent. Prepare 250 ml. of 1N H2SO4 (Concentration of commercial concentrated Molality : Molality is the number of moles of solute per one kilogram of H2SO4 = 36 N) solvent. It is denoted by m. Solution : number of moles of solute Concentration of commercial concentrated H2SO4 = 36 N m= weight of solvent in Kg. Concentation of dilute H2SO4 to be prepared = 1 N Higher concentration 36 Ex: 1m HCl means 36.5 g HCl in 1 Kg. water. Dilution factor = = = 36 Formality : It is the number of fomula weight in gm. dilssolved per lower concentration 1 litre of solution. When formula weight is equal to the molecular weight, i.e. 1 ml. of commercial concentrated H2SO4 diluted to 36 ml. makes 1N formality is same as Molarity. H2SO4 Mole Fraction : It is the ratio of number of moles of one of the Quantity of commercial concentrated H2SO4 required to make 250 ml. of components to the total number of moles of solute and solvent. It is 1N H2SO4 = denoted by ‘x’. If ‘n1’ is number of moles of solute, ‘n2’ is number 36 ml. -------- 1ml. of moles of solvent, ‘x1’ is mole fraction of solute and ‘x2’ is mole fraction 250 ml. --------- ? of solvent, 250 X 1 n1 n2 = 7 ml. x1 = and x2 = 36 n1 + n2 n1 + n2 Quantity of commercial concentrated H2SO4 required to make 250 ml. of x1 + x2 = 1. 1N H2SO4 = 7 ml. Conversion of mg% to meq/L : Method of Preparation : mg. per 100 ml. X 10 X valency Concentration in meq/L = 1. Measure accurately 7 ml. of commercial concentrated H2SO4. molecular weight 2. Add gradually to about 190 ml. of water, mix and cool. 3. Dilute to 250 ml. with distilled water. 4. Standardise and adjust normality if necessary. 166 167  SUMMARY 5. What is a weight in volume percentage solution? Give the general formula. Solutions are monophasic mixtures of two or more substances in which 6. Define weight in weight percentage solutions. Write general formula. solute is dissolved in solvent. Solutions are classified on the basis of states of 7. Define volume in weight percentage solutions. Write the general formula. matter of solute and solvent. Concentrations are expressed as percentage, molarity, normality, molality, mole fraction and formality. Percentage solutions are wieght 8. What is a volume in volume percentage solution? Give the general formula. in volume (w/v) solutions, weight in weight (w/w) solutions, volume in weight (v/ 9. Define molarity. w) solutions and volume in volume (v/v) solutions. 10. What is a molar solution? Molarity is number of moles of substance present in one litre solution. 11. Define mole. Normality is number of gram equivalents present in one litre of solution. Molality is number of moles present in one Kg. solvent. Mole fraction is ratio of number 12. What is meant by gram molecular weight? of moles of one of the components of the solution to the total number of moles of 13. Write molarities of commercially available a) Concentrated HCl b) solute and solvent. Formality is number of formula weight in gm dissolved in litre Concentrated H2SO4. of solution. 14. Mention specific gravities of commercially available a) Concentrated HCl b) Essay Questions Concentrated H2SO4. 1. Define solutions. Classify and exemplify solutions. 15. Mention the percentages by weight of actual content present in commercially 2. What are different methods of expressing concentration? Classify solutions available a) Concentrated HCl b) Concentrated H2SO4 on that basis. 16. Give the normalities of commercially available a) Concentrated HCl b) 3. Mention different types of percentage solutions. Write about weight in volume Concentrated H2SO4. percentage solutions and their preparation. 17. Mention the relation between molarity and normality. 4. What are the general applications of different types of percentage solutions? 18. What is molality? Write about volume in volume percentage solutions and their preparation. 5. Write about molar solutions and thier preparation. 19. What is molal solution? 6. Write about normal solutions and their preparation. 20. Define formality. Short Answer Questions 21. What is mole fraction? 1. Define solution. PROBLEMS 2. Define solute, solvent. 1. Prepare 75 ml. of 1% w/v glucose stock standard solution with 0.25% w/v aqueous benzoic acid solution as solvent. 3. Give examples of a) Solid in liquid solution b) Liquid in liquid solution. 2. Prepare 60 ml. of 0.1% w/v glucose working standard from 1% w/v glucose 4. What is percentage solution? stock standard solution. 168 169 3. Prepare 100 ml. of 70% v/v alcohol from absolute alcohol 4. Prepare 150 ml. of 80% v/v alcohol from absolute alcohol. 5. Prepare 60 ml. of 0.2M HCl. 6. Prepare 75 ml. of 0.5M HCl from commercial concentrated HCl. (Specific gravity of commercial concentrated HCl = 1.16, % age by weight = 36% w/w) 7. How do you prepare 300 ml. of 0.2 M H2SO4 from commercially available concentrated H2SO4? (Specific gravity of commercial concentrated H2SO4 = 1.84 % age by weight - 95% w/w) 8. Prepare 250 ml. of 0.2 M HCl by dilution from commercial concentrated HCl (molarity of commercial concentrated HCl = 12M) 9. Prepare 250 ml. of 0.1 N HCl. 10. Prepare 300 ml. of 0.5 N NaOH. 11. Prepare 500 ml. of N/10 HCl from commercial concentrated HCl. (Specific gravity of commercial concentrated HCl = 1.16 % age by weight - 36% w/w) 12. Prepare 500 ml. of 2/3 NH2SO4. (Specific gravity of commercial concentrated H2SO4 = 1.84, % age by weight = 95% w/w) 13. Prepare 250 ml. of 2/3 NH2SO4 from commercial concentrated H2SO4 by dilution (concentration of commercial concentrated H2SO4 = 36 N) 170