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1 PRT - : SOLUTION. SOLUTION solution is a homogeneous mixture of a solute, the substance that dissolves and a solvent, the substance in which the solute dissolves. (a) The component of solution which is in lesser amount (Which is dissolved) is called solute. (b) The component of solution in which solute is dissolved is called solvent. Solution Solute + Solvent. CONCENTRTION OF SOLUTION It is calculated by following two methods (a) Weight % : Weight of solute per gram of solution. (b) Volume % : (i) Weight of solute per ml of solution (ii) Volume of solute per ml of solution (c) Concentration of a solution expressed in following terms.. MOLRITY (M) It is the number of moles of solute in one litre of solution No. of moles of solute Molarity Volume of solution ( litre) (a) Molarity is expressed by putting a suffix M after a number, say X. It means if concentration of a solution is given to be XM, it means X moles of solute are there per litre of solution. (b) Some times amount of solute is given in grams. So, amount ( gram) Moles of solute mol. wt. in ( gram) (c) We Should be careful about unit of volume taken. We have to use volume in litre in the formula for molarity. So, if volume is given in ml (mili litre) convert it into litre asvolume in litre volume in ml 3 (d) Some times we get confused when volume is given in cm 3. cm 3 is nothing but a milliliter. So, volume in litre volume (cm 3 ) 3 (e) Unit of molarity is mole L. (f) Millimoles M V(ml) wt. mol. wt. (g) Strength of solution wt. of solute volume of solution ( ml) (h) Molarity is a temperature dependent unit.. MOLLITY (m) Molarity temp. It is the number of moles of solute per kilogram of solvent. Molality (a) Unit of molality is mole kg moles of solute mass of solvent ( kg) (b) Sometimes mass of solution is given instead of solvent, so subtract the mass of solute to get mass of solvent (c) Represented by a suffix m after a number x. xm means x moles of solute are there per kg of solvent. (d) Molality does not change with increase of temperature. Where, m Molality M Molarity (e) Relation between molarity and molalitym M d MM d density of solution in gm/litre M Molecular wt. of solute Derivation : M m wt. of solvent Volume of solution M m W V( litre) Since,wt. of solvent wt. of solution wt. of solute Therefore, M m V d MM Vd MM V MV m Vd MM.3 MOLE FRCTION (X) (a) Mole fraction of a component in solution is equal to the ratio of number of moles of that component to the total number of moles of all the components in the solution. Mole fraction of component is represented by x. - Free NEET & IIT Study Meterial & Papers Page -

2 (b) Let, there be two components (solvent) & (solute) X n n n and X n n n where, X Mole fraction of solvent and X Mole fraction of solute. Here, n w M (wt. of in grams/mol.wt. of, x + x ) (c) It is temperature independent unit. (d) Relation between mole fraction and molality - Mole fraction of solvent n X n n Mole fraction of solute X n n n X n X n On multiplying in both side X X n n X X.4 MSS FRCTION : m M Ratio of mass of component to the total mass of components mass fraction of w w w where, w weight of, and w weight of..5 MOLE PERCENT : number of moles of a component in moles Mole percent mass fraction.6 PRTS PER MILLION (PPM) : (a) amount of component in mg in kg of solution. ppm ppm Mass of solute Mass of solution 6 wt. of wt. of solute solute wt. of solvent 6 (b) Generally, it is used f or v ery-v ery little concentrations..7 NORMLITY (N) : (a) Number of gram equivalents of a solute per litre of solution. Its unit is gram equivalents per litre. Normality (b) Normality gram equivalent of solute Volume of solution (litres) Wt. of solute equivalent wt. of solute Volume of solution ( ml) (c) N W where, W wt. of solute and E V E equivalent wt. of solute (d) Equivalent N V (litre) wt. of solute equivalent wt. of solute (e) Milliequivalent N V (ml) wt. of solute equivalent wt. of solute (f) Strength of solution (in gm/lit) Normality (N) Eq. wt. of solute (E) ( specific gravity or density ) wt.% of solute equivalent wt. of solute (g) Normality is a temperature dependent quantity. (h) Concentration in grams L Eq.wt. Normality Mol. wt. x Molarity Examples based on Ex. Solution and concentration of solution Find the normality of H SO 4 having 5 milli equivalents in litre. ().5 ().5 (C).5 (D).5 (ns. ) N V(In L) Equivalent 5 N N.5 Ex. 6.9 M solution of KOH in water has 3% by weight of KOH. Calculate density of solution. ().88 g ml ().88 g ml (C) 4.88 g ml (D).488 g ml (ns. ) - Free NEET & IIT Study Meterial & Papers Page -

3 Ex.3 Ex.4 KOH solution is 3% by weight. wt. of KOH 3 g and Wt. of solution g Volume of solution d Molarity 6.9 F G HG.88 g ml 3 56 I J djk.8g of a metal required 3.5 ml of N HCl to dissolve it. What is equivalent weight for metal? () 65 () 75 (C) 56 (D) 9 (ns. C) Meq. of metal Meq. of HCl or. 8 E 3.5 E 56 Calculate the amount of oxalic acid (H C O 4. H O) required to obtain 5 ml of semimolar solution. () 5.75 g ().575 g (C) 57.5 g (D) None (ns. ) Molarity of solution.5 M Volume of solution 5 ml milli mole of oxalic acid M V (ml) 5 or w M.5 5 w g Ex.5 One g of calcium was burnt in excess of O and the oxide was dissolved in water to make up one litre solution. Calculate normality of alkaline solution ().5 ().5 (C).5 (D).5 (ns. C) Ca + ½O CaO Equivalent taken / excess Equivalent after / ( Eq. w/e) reaction Ex.6 N CaO F HG Eq..5 N V in litre W hat is mole fraction in its one molal aqueous solution- ().8 ().8 (C).8 (D) None (ns. ) Mole fraction Ex.7 n n n n and n To neutralize ml of a(oh) solution, we have to add ml. of. M H SO 4 solution. What is the normality of a(oh)? ().5 (). (C). (D). (ns. D) (. M H SO 4. N H SO 4 ) Ex.8 [H SO 4 ] N V N V [a(oh) ]. N. N How much volume of M H SO 4 is needed to neutralize ml of M NaOH solution? () 5 ml () 5 ml (C) ml (D) ml (ns. ) Normality of H SO 4 N N V N V N N V 5ml V I K J - Free NEET & IIT Study Meterial & Papers Page - 3

4 PRT - : COLLIGTIVE PROPERTIES. COLLIGTIVE PROPERTIES Certain properties of dilute solutions containing nonvolatile solute do not depend upon the nature of the solute dissolved but depend only upon the concentration. i.e the number of the particles of the solute present in the solution. Such properties are called colligative properties. The four well known examples of the colligative properties are - (a) Lowering of vapour pressure of the solvent (b) Osmotic pressure of the solution (c) Elevation in boiling point of the solvent (d) Depression in freezing point of the solvent. VPOUR PRESSURE OF LIQUID It is the pressure that its vapours exert when in equilibrium with the liquid at a given temperature. It depends upon the following factors : (i) Nature of solvent (ii) Temperature (iii) % age purity of liquid or surface area of liquid In other words a pure liquid always has a vapour pressure greater than its solution. More is the force of attraction, lower is the vapour pressure and vice-versa. 3. ROULT S LW (a) Raoult proposed a law which states that at a given temperature, the vapour pressure of a solvent in a solution containing non-volatile solute is directly proportional to its mole fraction. Mathematically, p p x solvent Vapour Pressure P X X P p X + p x P p x Mole Fraction P pº X X X P Vapour Pressure diagram for an ideal solution (b) In the case of binary solutions of two volatile liquids, Raoult's law states that at a given temperature, the partial vapour pressure of any component of the solution is equal to the product of the vapour pressure of the pure component and its mole fraction in the solution i.e. p p X and p p X (c) The total vapour pressure P of such a solution containing two components and is, P p + p p X + p X ( - X ) p + p X (p - p ) X + p where, p and p are vapour pressures (of pure component) and X and X are mole fractions of components and respectively. Plot of P should be a straight line which is true for ideal solution. Thus the addition of a solute may raise or lower the vapour pressure of solvent depending upon which one is more volatile. (d) Raoult s Law mathematically expressed as P P P S n n N where,p Vapour pressure of pure solvent P S Vapour pressure of solution n moles of non-volatile solute N moles of solvent If the solution is very dilute, then n << N So, or P P P S P P P S n N w / m W / M where,w wt. of solute dissolved in grams w wt. of solvent in grams m molecular mass of solute M molecular weight of solvent P P P S w. m W. M... (i) (ii)... (iii) This expression (iii) can be used for calculating the molecular weight of solutes. 3. LIMITTIONS OF ROULT S LW (a) s described earlier, Raoult s law is applicable only to very dilute solutions. (b) Raoult s law is applicable to solutions containing non-volatile solute only. (c) Raoult s law is not applicable to solutes which dissociate or associate in the particular solution - Free NEET & IIT Study Meterial & Papers Page - 4

5 Ex. Raoult s law Examples based on Ex.9 t 3 K, the vapour pressure of an ideal solution containing one mole of and 3 moles of is 55 mm of Hg. t the same temperature, if one mole of is added to this solution, the vapour pressure of solution increases by mm of Hg. Calculate the vapour pressure of and in their pure state. () 4 mm, 6 mm (C) mm, 3 mm () 6 mm, 4 mm (D) 3 mm, mm Initially, P M Pº. X + Pº. X 55 Pº 3 or Pº + 3Pº (ns. ) F I HG K J F 3 + Pº HG When mole of is further added to it 3 P M Pº. X + Pº. X 56 Pº 4 or Pº + 4Pº 8 y (i) and (ii) Pº 4 mm ; Pº 6 mm F I HG K J F 4 + Pº HG 4 Ex. The vapour pressure of pure liquid at 3ºC is torr. The vapour pressure of this liquid in solution with liquid is 7 torr. Calculate the mole fraction of in solution if the mixture obeys Raoult s law. I K J I K J ().6 ().9 (C).3 (D).6 (ns. D) Given is vapour pressure of pure component, Pº torr Partial vapour pressure of, P 7 torr Suppose, its mole fraction in solution is x, then according to Raoult s law or x P Pº. x 7 x 7.6 n aqueous solution containing 8% by weight of liquid (mol. wt. 4) has a vapour pressure of 6 mm at 37ºC. Find the vapour pressure of the pure liquid (the vapour pressure of water at 37ºC is 5 mm). () mm () mm (C) mm (D) None (ns. C) We know that for a mixture of two miscible liquids, P total Pº.x + Pº.x where x and x are mole fractions of liquids and, and Pº and Pº are their vapour pressure in pure state. Moles of liquid 8 4. Moles of water (liquid ) Total number of moles in the solution Moles fraction of Mole fraction of water () P total Vapour pressure of the solution 6 mm (given) So, 6 Pº a or Pº or Pº 4. IDEL SOLUTIONS mm These are the solutions in which solute-solute and solvent-solvent interactions are almost similar to solute-solvent interactions. Ideal solutions obey Raoult's law for all range of concentrations and temperature. H mix V mix eg. Hexane + Heptane, ethyl chloride + ethyl bromide,chlorobenzene + romobenzene etc. 5. NON-IDEL SOLUTIONS (a) These are the solutions in which solute-solvent interactions are different than solute-solute and solvent-solvent interactions. The non-ideal solutions do not obey Raoult's law for all concentrations H mix V mix - Free NEET & IIT Study Meterial & Papers Page - 5

6 (b) These non-ideal solutions show two types of deviations from the ideal behaviour. (i) If V mix > and H mix >, then non-ideal solutions show positive deviations. (ii) If V mix < and H mix <, then non-ideal solutions show negative deviations 6. ZEOTROPIC MIXTURES 5. TYPES OF NON-IDEL SOLUTIONS (a) Non-ideal solutions showing positive deviations : In such a case, the observed vapour pressure of each component and the total vapour pressure are greater than predicted by Raoult's law i.e. Vapour Pressure p > p X, p > p x, P > p + p This is because the new interactions are weaker than those in the pure components. eg. cetone + Ethyl alcohol, Water + Ethyl alcohol, CCl 4 + CHCl 3 Ethanol + CHCl 3 Positive deviation (solid lines) from Raoult's law (dotted lines). (b) Non-ideal solutions showing negative deviations : In such a case the observed vapour pressure of each component and the total vapour pressure are less than predicted by Raoult's law i.e. Vapour Pressure P P X X P p X + p x Mole Fraction P p X + p x Mole Fraction X X P P p < p X, p < P X, P < p + p This is because the new interactions are stronger than those in the pure components. eg. cetone + niline, HCl + water, HNO 3 + water, water + H SO 4 etc Negative deviations (solid lines) from Raoult's law (dotted lines) Note : S mix > for ideal solutions +ve the deviation as well as ve deviation solution, since no. of total particles increase on mixing. zeotropic mixtures of two liquids which boil at a constant temperature and can be distilled unchanged in their composition. They are formed by non-ideal solutions. 6. TYPES OF ZEOTROPIC MIXTURES (a) "Minimum boiling azeotropes" are the mixture of two liquids, whose boiling points is less than either of the two pure components. They are formed by non-ideal solution showing positive deviation. eg. ethanol (95.5%) + water (4.5%) mixture boiling at 35.5 K. (b) "Maximum boiling azeotropes" are the mixtures of two liquids, whose boiling points are more than either of the two components. They are formed by non-ideal solution showing negative deviation. eg. HNO 3 (68%) + water (3%) mixture boiling at K 7. OSMOTIC PRESSURE ( ) (a) Osmotic pressure may be defined as the excess pressure which must be applied to a solution in order to prevent flow of solvent into the solution through the semipermeable membrane. Osmotic pressure may also be defined in several other ways. (b) Osmotic pressure is the excess pressure which must be applied to a given solution in order to increase its vapour pressure until it becomes equal to that of the solution (c) Osmotic pressure is the negative pressure which must be applied to (i.e. the pressure which must be withdrawn from) the pure solvent in order to decrease its vapour pressure until it becomes equal to that of the solution (d) Osmotic pressure is the hydrostatic pressure produced when a solution is separated from the solvent by a semipermeable membrane. 7. MTHEMTICL EXPRESSION crt c concentration (mol.l ) R Gas constant (mol.l ) T bsolute temperature (mol.l ) - Free NEET & IIT Study Meterial & Papers Page - 6

7 8. COLLIGTIVE PROPERTIES OF DILUTE SOLUTIONS Here, refers to solvent, refers to solute * Molal elevation constant (K b ) (a) dilute solution is one in which the amount of the solute is very small in comparison to the amount of the solvent. (b) Dilute solutions containing non-volatile solute exhibit some special properties which depend only upon the number of solute particles present in the solution irrespective of their nature. These properties are termed as colligative properties. (c) The colligative properties are- (i) Relative lowering of vapour pressure (ii) Elevation in boiling point (iii) Depression in freezing point (iv) Osmotic pressure 8. EXPRESSION FOR DIFFERENT COLLIGTIVE PROPERTIES (a) Osmotic pressure ( ) n V RT CRT when w gram of solute are dissolved in V litres of solutions and M is the molar mass of the solute, then wrt MV W n M when height is involved hdg (h height, d density, g gravitational acceleration) For isotonic or isosmotic solutions or n V n V W MW V MV [ ] (b) Relative lowering in vapour pressure : p p p X n n N [n moles of solute, N moles of solvent] (c) Elevation in boiling point : T b K b x m (d) Depression in freezing point : T f K f x m K b W M W K f W M W K b RT b [l v latent heat of vapourisation] * Molal depression constant (K f ) Examples based on Ex. K f RT f f v [l f latent heat of fusion] Colligative properties of dilute solutions What will be the temperature at which a solution containing 6 g of glucose per g water will boil, if molal elevation constant for water is.5 / g. ().73ºC ().73ºC (C).73ºC (D) None (ns. ) w 6g, W g, Mol. wt. of glucose 8 Ex.3 T b Kb w m W ºC Hence boiling point of solution b.p. of water + T b ºC. If latent heat of fusion of ice is 8 cals per g at ºC, calculate molal depression constant for water. () 8.63 () 86.3 (C).863 (D).863 (ns. C) RT f K f v Here R cals, T f K, f 8 cals Ex.4 K f Calculate the molal elevation constant of water evaporates at ºC with the absorption of 536 calories per gm (R cals). ().59ºC ().59ºC (C).59ºC (D).59ºC (ns. ) - Free NEET & IIT Study Meterial & Papers Page - 7

8 Molal elevation constant of the solvent. K b RT b? v ºC Ex.5 The vapour pressure of CCl 4 (density.58 g cm 3 ) at 3ºC is 43 mm..5 g of a non-volatile solute of molecular weight 65 is dissolved in ml of CCl 4. Calculate the vapour pressure of the solution- () 4.93 mm () 4.93 mm (C).493 mm (D) None (ns. ) Here w.5 g, W g (since d W / V), m 65, or or M of CCl p º p pº 9. REVERSE OSMOSIS 43 p p 4.93 mm wm mw If a pressure higher than osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through the semipermeable membrane. Since here the flow of solvent is in the reverse direction to that observed in the usual osmosis, the process is called reverse osmosis.. ISOTONIC SOLUTIONS (a) pair of solutions having the same osmotic pressure are known as isosmotic solutions. If two such solutions are separated by a semipermeable membrane there will be no transference of solvent from one solution to the other. Isosmotic solutions separated by a semipermeable membrane are known as isotonic or Iso-osmotic solutions. (b) Isotonic solutions hav e the same molar concentration. eg..85% NaCl solutions is found to be isotonic with blood. (c) solution having lower or higher osmotic pressure than the other is said to be hypotonic or hypertonic respectively in respect to other solution. (d) When placed in hypotonic solutions, cells swell and burst (haemolysis) (e) When placed in hypertonic solutions, cells contract in size (plasmolysis). When excess of fertilizers (like urea) are applied, plasmolysis takes place and plants dry up (wilt).. COLLIGTIVE PROPERTIES OF ELECTROLYTES The colligative properties of solutions, viz, lowering of vapour pressure, osmotic pressure, elevation in b.p. and depression in freezing point, depend solely on the total number of solute particles present in solution. Since the electrolytes ionise and give more than one particle per formula unit in solution, the colligative effect of an electrolyte solution is always greater than that of a nonelectrolyte of the same molar concentration. (a) Colligative properties Number of particles Number of molecules (in case of nonelectrolytes) Number of ions (In case of electrolytes) Number of moles of solute Mole fraction of solute (b) For different solutes of same molar concentration, the magnitude of the colligative properties is more for that solution which gives more number of particles on ionisation. (c) For different solutions of same molar concentration of different nonelectrolytes solutes, the magnitude of the colligative properties will be same for all. (d) For different molar concentrations of the same solute, the magnitude of colligative properties is more for the more concentrated solution. (e) For solutions of different solutes but of same percent strength, the magnitude of colligative property is more for the solute with least molecular weight. (f) For solutions of different solutes of the same percent strength, the magnitude of colligative property is more for that solute which gives more number of particles which can be known by the knowledge of molecular weight and its ionisation behaviour. eg. mong the.m solutions of urea, NaCl, acl, Na 3 PO 4 and l (SO 4 ) 3 solutions (i) Vapour pressure and freezing point will be lowest while b.p. will be highest for l (SO 4 ) 3 solution (ii) The values of the four colligative properties will be highest for l (SO 4 ) 3 solution eg. mong % solution of urea, glucose and sucrose (i) Vapour pressure and freezing point area lowest while boiling point is highest for urea solution (ii) The four colligative properties are highest for urea solution eg. mong.m glucose,.5m urea and.m sucrose solutions (i) Vapour pressure and freezing point is lowest, while boiling point is highest for sucrose solution (ii) The four colligative properties are highest for sucrose solution. - Free NEET & IIT Study Meterial & Papers Page - 8

9 . VN T HOFF FCTOR Certain solutes which undergo dissociation or association in solutions are found to show abnormal molecular mass. Thus, in order to know about the extent of association or dissociation of solutes in solution Vant Hoff introduced a factor (i). It is defined as the ratio of the normal mass to the observed molecular mass of the solute i.e. i i i i Normal molar mass Observed molar mass ; Observed colligative property Normal colligative property Observed osmotic pressure Normal osmotic pressure ctual number of paticles No. of particles for no ionisation. Van t Hoff factor and degree of association : If a solute forms associated molecules n and is the degree of association then, n n Initial mole mole ssociated change mole + /n mole Equilibrium mole (- ) mole /n mole Total number of moles after association mole of ( ) + ( /n) mole Van't Hoff factor (i) ( ) ( n ) mole Number of moles after association Normal number of mole taken ( )( / n ) n ; + n degree of association n ( i) n. Van't Hoff factor and degree of dissociation : If a molecule of solute on dissociation gives n ions and is the degree of dissociation then, () n n Initial mole mole dissociation change mole + n mole Equilibrium mole ( ) mole n mole Total number of moles after dissociation of mole of [( - ) + n ] mole + (n - ) mole Van't Hoff factor (i) Number of moles after dissociation Number of moles taken(normal) (n ) + n degree of dissociation Examples based on Ex.6 Van t Hoff factor + (n ) i ( ) n 7.6g Kr in 5 ml solution was found to have an osmotic pressure of.84 atm at 7ºC. Calculate degree of ionisation and Van t Hoff factor. () 43.4% () 45.4% (C) 4.34% (D) 4.45% (ns. ) exp.84 atm lso N V w m ST N N.58 atm (i) exp N + or i Thus, Van t Hoff factor i.434 and.434 or 43.4% Ex.7 The freezing point of a solution containing.g of acetic acid in. g benzene is lowered by.45ºc. Calculate the degree of association of acetic acid in benzene. ssume acetic acid dimerizes in benzene. K f for benzene 5. K mol kg. () 49.5 % () 94.5% (C) 85.5% (D) 58.5% (ns. ) Given, w. g, W g, T.45ºC T K w m W - Free NEET & IIT Study Meterial & Papers Page - 9

10 or.45 m(observed) m Now for CH 3 COOH (CH 3 COOH) efore association fter association / Where is degree of association or m m normal observed / or.945 or 94.5 % + / Ex.8.6% aqueous solution of KCl (mol. wt. 74.5) freezes at.ºc. Calculate the vant Hoff factor and degree of dissociation of KCl, K f for water.86ºc () 9.7 % ().97 % (C) 5.56 % (D) 55.6% (ns. ) Observed mol. wt. of KCl Vant Hoff factor, Kf w T W i f Calculated mol wt. Observed mol wt Let x be the degree of dissociation of dissociation of KCl i.e., KCl K + + Cl x x x Number of particles after dissociation x + x + x + x LIQUID STTE Note : The syllabus of Liquid state includes properties of liquids, vapour pressure, surface tension and viscosity which has been discussed in detail in the chapter of colligative properties (in chemistry) and in the chapter of surface tension and viscosity (in physics). Therefore, the theory and problems are not seperately given in this sheet. Thus, Normal mol. wt. Observed mol. wt x or x.97 or 9.7% - Free NEET & IIT Study Meterial & Papers Page -

11 Ex. Ex. What is the mass of sodium bromate solution necessary to prepare 85.5cm 3 of.67 N solution. When the half cell reaction is- ro H + + 6e r - + 3H O ().46 g ().46g (C).346 g (D).446 g (ns. 4) Since 6e are involved in the reaction, we have Molarity Normality M Molar mass of NarO 3 5 g mol - Mass of NarO 3 needed to prepare required F I HG K J.446 gm. solution The density of a solution containing 3% by mass of sulphuric acid is.9 g/ml. Calculate the molarity and normality of the solution- ().445 M () 4.45 M (C) 44.5 M (D).445 M (ns. ) Volume of gram of the solution d Ex.3. 9 ml. 9 litre. 9 litre Number of moles of H SO 4 in gram of the solution 3 98 Molarity No. of moles of HSO4 Volume of solution in litre M Calculate the molarity of pure water (d g / L) () 555 M () 5.55 M (C) 55.5 M (D) None (ns. C) Consider ml of water Mass of ml of water gram Number of moles of water 8 SOLVED EXMPLES 55.5 Ex.4 Ex.5 Molarity No. of moles of water Volume in litre M Calculate the quantity of sodium carbonate (anhydrous) required to prepare 5 ml solution- ().65 gram () 4.95 gram (C) 6.5 gram (D) None (ns. ) We know that Molarity where; W M V W Mass of Na CO 3 in gram M Molecular mass of Na CO 3 in grams 6 V Volume of solution in litres 5.5 Molarity Hence, or W W gram ml of. M KMnO 4 was required to completely oxidise ml of oxalic acid solution. What is the molarity of the oxalic acid solution? (). M ().4 M (C). M (D) 4. M (ns. ) The concerned reaction is KMnO 4 + 3H SO 4 + 5(COOH) K SO 4 + MnSO 4 + CO + 8H O [For KMnO 4 ] [For oxalic acid] V ml V ml M. M M? n n M V n. M V n M M.. M - Free NEET & IIT Study Meterial & Papers Page -

12 Ex.6 Find the molality of H SO 4 solution whose specific gravity is.98 g ml - and 95% by volume H SO 4 () 7.4 () 8.4 (C) 9.4 (D).4 (ns. C) H SO 4 is 95% by volume wt. of H SO 4 95g Vol of solution ml Moles of H SO solution x g, and weight of Weight of water g Molality Hence molality of H SO 4 solution is 9.4 Ex.7 Calculate molality of litre solution of 93% H SO 4 by volume. The density of solution is.84 gm ml Ex.8 () 9.4 ().4 (C).4 (D).4 (ns. ) Given H SO 4 is 93% by volume wt. of H SO 4 93g Volume of solution ml weight of solution.84 gm 84 gm wt. of water gm Molality Moles wt. of water in kg Suppose 5gm of CH 3 COOH is dissolved in one litre of Ethanol. ssume no reaction between them. Calculate molality of resulting solution if density of Ethanol is.789 gm/ml. ().856 ().956 (C).56 (D).56 (ns. C) Wt. of CH 3 COOH dissolved 5g Eq. of CH 3 COOH dissolved 5 6 Volume of ethanol litre ml. Weight of ethanol g Ex.9 Ex. Molality of solution Moles of solute Wt. of solvent in kg Calculate the molarity and normality of a solution containing.5 gm of NaOH dissolved in 5 ml. solution- ().5 M,.5 N ().5 M,.5 N (C).5 M,.5 N (D).5 M,.5 N (ns. ) Wt. of NaOH dissolved.5 gm Vol. of NaOH solution 5 ml Calculation of molarity.5 g of NaOH. 5 4 moles of NaOH [ Mol. wt of NaOH 4].5 moles Thus 5 ml of the solution contain NaOH.5 moles ml of the solution contain M Hence molarity of the solution.5 M Calculation of normality Since NaOH is monoacidic ; Eq. wt. of NaOH Mol. wt. of NaOH 4.5 gm of NaOH gm equivalents gm equivalents Thus 5 ml of the solution contain NaOH.5 gm equ. ml of the solution contain Hence normality of the solution.5 N Calculate the molality and mole fraction of the solute in aqueous solution containing 3. gm of urea per 5 gm of water (Mol. wt. of urea 6). (). m,.357 ().4 m,.357 (C).5 m,.357 (D).7m,.357 (ns. ) - Free NEET & IIT Study Meterial & Papers Page -

13 Wt. of solute (urea) dissolved 3. gm Wt. of the solvent (water) 5 gm Mol. wt. of the solute 6 Ex. 3. gm of the solute 3. 6 moles.5 moles Thus 5 gm of the solvent contain.5 moles of solute gm of the solvent contain moles Hence molality of the solution. m In short, Molality No. of moles of solute/ g of solvent Molality 3 / 6 5. m Calculation of mole fraction 3. gm of solute 3 / 6 moles.5 moles 5 gm of water 5 8 moles 3.94 moles Mole fraction of the solute solution has 5% of water, 5% ethanol and 5% acetic acid by mass. Calculate the mole fraction of each component. ().5,.3,.9 ().9,.3,.5 (C).3,.9,.5 (D).5,.9,.3 (ns. 4) Since 8 g of water mole 5 g of water 5.38 mole 8 Similarly, 46 g of ethanol mole 5 g of ethanol moles gain, 6 g of acetic acid mole 5 g of acetic acid 5 6 Mole fraction of water.83 mole Similarly, Mole fraction of ethanol Mole fraction of acetic acid Ex. 5 gram of methyl alcohol is dissolved in 35 gram of water. What is the mass percentage of methyl alcohol in solution? () 3% () 5% (C) 7% (D) 75% (ns. ) Total mass of solution (5 + 35) gram 5 gram mass percentage of methyl alcohol Ex.3 Mass of methyl alcohol Mass of solution 5 5 3% Calculate the masses of cane sugar and water required to prepare 5 gram of 5% cane sugar solution- () 87.5 gram, 6.5 gram () 6.5 gram, 87.5 gram (C) 6.5 gram, 87.5 gram (D) None of these (ns. ) Mass percentage of sugar 5 We know that Mass percentage So, 5 Mass of solute Mass of solution Mass of cane sugar. 5 or Mass of cane sugar gram Mass of water (5-6.5) 87.5 gram Ex.4 Calculate normality of the mixture obtained by mixing ml of.n HCl and 5ml of.5n NaOH solution. ().467 N ().367 N (C).67 N (D).67 N (ns. D) Meq. of HCl. Meq. of NaOH HCl and NaOH neutralize each other with equal eq. - Free NEET & IIT Study Meterial & Papers Page - 3

14 Meq. of NaOH left Volume of new solution ml. N NaOH left N Hence normality of the mixture obtained is.67 N Ex.5 3 ml. M HCl and ml of.3m H SO 4 are mixed. Calculate the normality of the resulting mixture- ().84 N ().84 N (C).4 N (D).84 N (ns.) For HCl For H SO 4 V 3 ml V ml Ex.6 Ex.7 N M x asicity N M x asicity. x..3 x.6 Normality of the mixture, N V N V N V V N 5 In what ratio should a 6.5 N HNO 3 be diluted with water to get 3.5 N HNO 3? () 6 : 7 () 7 : 6 (C) 5 : 6 (D) 6 : 5 (ns. ) Using cross wise rule x y The dilution ratio is 3.5 : 3 or 7:6 i.e 6 volumes of water should be added to 7 volumes of 6.5 N solution to get 3.5N HNO 3 solution. Calculate the amount of each in the following solutions- (i) 5 ml of N 7 H SO 4 (ii) 5 ml of.m NaHCO 3 (iii) 4 ml of N Na CO 3 (iv) 5 g of m KOH. Ex.8 () 5g,.g, 4.g,.5g ().5g, 4.g,.g, 5g (C).5g,.g, 5g, 4.g (D) 4.g,.g,.5g, 5g (ns. ) (i) Eq. wt. of H SO 4 Mol. wt. asicity mount of H SO 4 per litre (strength) Normality Eq. wt. 7 mount in 5 ml 7 5 (ii) Molecular wt. of 49 7 g/ litre.5 g NaHCO mount of NaHCO 3 required to produce c.c. of one molar solution 84 g mount present per litre in. M solution g mount present in 5 c.c (iii) Equivalent weight of Na CO 3 4. g Mol. wt. No. of positive valencies 6 53 mount of Na CO 3 Normality Eq. wt g/litre mount present in 4 c.c g (iv) We know that molal solution of a substance contains g of solvent. Wt. of KOH in 5 g of m KOH solution 5 5 g How many kilograms of wet NaOH containing % water are required to prepare 6 litres of.5 N solution? ().36 kg ().5 kg (C).4 gm (D) 3.6 kg (ns. ) One litre of.5 N NaOH contains.5 4g g. kg 6 litres of.5 N NaOH contain. 6 kg. kg NaOH - Free NEET & IIT Study Meterial & Papers Page - 4

15 Since the given NaOH contains % water, the amount of pure NaOH in kg of the given NaOH 88 kg Ex.9 Thus 88 kg of pure NaOH is present in kg wet NaOH. kg of pure NaOH is present in kg wet NaOH Two liquids and form ideal solution. t 3 K, the vapour pressure of a solution containing mole of and 3 mole of is 55 mmhg. t the same temperature, if one mole of is added to this solution, the vapour pressure of the solution increases by mm of Hg. Determine the vapour pressure of and in their pure states. () 4 mm, 6 mm () 6 mm, 4mm (C) 3 mm, mm (D) mm, 3mm (ns. ) Let the vapour pressure of pure be p ; and the vapour pressure of pure be p. Total v apour pressure of solution ( mole + 3 moles ) X.p + X.p [X is mole fraction of and X is mole fraction of ] 55 4 p p or p + 3p... (i) Total v apour pressure of solution ( mole + 4 moles of ) 5 p p 56 5 p p or 8 p + 4p... (ii) Solving equation (i) and (ii), p 6 mm of Hg vapour pressure of pure p 4 mm of Hg vapour pressure of pure Ex. Calculate the vapour pressure of a solution at C containing 3g of cane sugar in 33g of water. (t wt. C, H, O 6) () 76 mm () mm (C) mm (D) None (ns. ) Vapour pressure of pure water (solvent) at C, p 76 mm. Vapour pressure of solution, p? Ex. Ex. Wt. of solvent, W 33g Wt. of solute, w 3g Mol. wt. of water (H O), M 8 Mol. wt. of sugar (C H O ), m x + x + x 6 34 ccording to Raoult s law, p p? p p p p 76 wm Wm w x M m x W 3 x 8 34 x 33 x p x 76 ( p for H O 76 mm) mm n aqueous solution containing 8% by mass of a liquid (mol. mass 4) has a vapour pressure of 6 mm at 37 C. Find the vapour pressure of the pure liquid. (The vapour pressure of water at 37 C is 5 mm). () 36 mm () 5 mm (C) 6 mm (D) None (ns. ) For two miscible liquids, P total mol. fraction x p mol. fraction x p 8 No. of moles of 4. Liquid is water. Its mass is ( - 8), i.e. 7 No. of moles of Total number of moles Given P total 6 mm 5 mm p So, x p x 5 p 75. x mm 36 mm Osmotic pressure of a sugar solution at 4 C is.5 atmospheres. Determine the concentration of the solution in gm mole per litre. ().8 moles/litre ().8 moles/litre (C).5 moles/litre (D).87 moles/litre - Free NEET & IIT Study Meterial & Papers Page - 5

16 (ns. C) Here it is given that.5 atm, T K, S.8 lit. atm. deg - mol -, C? We know that CST Ex.3 Ex.4 or C? ST x 97.5 moles/litre Twenty grams of a substance were dissolved in 5 ml. of water and the osmotic pressure of the solution was found to be 6 mm of mercury at 5 C. Determine the molecular weight of the substance- () () 98 (C) (D) None of these (ns. ) Here it is given that w gm ; V 5 ml. 5.5 litre 6 mm 6 76 atm ; T m? ccording to Van t Hoff equation, V nst V w m ST m wst?v 98 x.8 x 88 x 76 6 x.5 lood plasma has the following composition (milli-equivalents per litre). Calculate its osmotic pressure at 37 C. Na + 38, Ca + 5., K + 4.5, Mg +., Cl 5, HCO 3 5, PO 3 4., SO , Proteins 6, Others. () 7.47 atm () 7.3 atm (C) 7.9 atm (D) 7.4 atm (ns. ) Since f or calculating osmotic pressure we require millimoles/litre therefore Na + 38 Ca , K + 4.5, Ex.5 Mg +.., Cl 5, HCO 3 4,PO , 3 SO , Proteins 6, others. Total 94.8 millimoles/litre.94 moles/litre Now since CST.94 x.8 x atm g of a substance dissolved in 5g of solvent boiled at a temperature higher by.6 C than that of the pure solvent. Calculate the molecular weight of the substance. Molal elevation constant for the solvent is.6 C. () 6 () (C) 78 (D) None of these (ns. ) Here it is given that w.5 g, T b.6 C W 5g K b.6 C m? Substituting values in the expression, m x K b x w? T x W x 6. x. 5 m. 6 x 5 Ex.6 solution of.45 gm of urea (mol. wt 6) in.5 g of water showed.7 C of elevation in boiling point. Calculate the molal elevation constant of water- ().7ºC ().45ºC (C).5ºC (D).3ºC (ns. C) Wt. of solute, w.45 g Wt. of solvent, W.5 g Mol. wt of solute, m 6 Molal elevation constant K b? oiling point elevation, T b.7 C Substituting these values in the equation- K b m x W x? T x w b b 6 x. 5 x 7..5 x. 45 C - Free NEET & IIT Study Meterial & Papers Page - 6

17 Ex.7 Calculate the boiling point of a solution Ex.8 containing.45g of camphor (mol. wt. 5) dissolved in 35.4g of acetone (b.p C); K b per gm of acetone is 7. C. () ºC () 5.4ºC (C) 56.46ºC (D) 5.464ºC Here it is given that w.45 g, W 35.4, m 5, Kb 7. per gm Now we know that T b x K b x w m x W (Note that this is expression when K b is given per g of the solvent) Substituting the v alues in the abov e expression. T b x 7. x x Now we know that.46 C.P. of solution (T) -.P. of solvent (T ) T.P. of solution (T).P. of solvent (T ) + T Hence.P. of solution C The freezing point of. molal K SO 4 is.ºc. Calculate Van t Haff factor and percentage degree of dissociation of K SO 4. K f for water is.86º () 97.5 () 9.75 (C) 5.5 (D) (ns. ) T f freezing point of water freezing point of solution º C (.º C).º We know that, T f i K f m. i.86.. i ut we know i + (n ).95 + (3 ) Van t Haff factor (i).95 Degree of dissociation.975 Percentage degree of dissociation Free NEET & IIT Study Meterial & Papers Page - 7