Chapter-1 BASIC CONCEPT OF CHEMISTRY

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1 Chapter-1 ASIC CONCEPT OF CHEMISTRY Law of conservation of mass : In all physical and chemical changes, the total mass of the reactants is equal to that of the products or Matter can neither be created nor destroyed. Law of constant composition or definite proportions : A chemical compound is always found to be made up of the same elements combined together in the same fixed ratio by weight. Law of multiple proportions : When two elements combine together to form two or more chemical compounds, then the weights of one of the elements which combine with a fixed weight of the other bear a simple ratio to one another. Law of Reciprocal Proportions : The ratio of the weights of two elements A and which combine with a fixed weight of the third element C is either the same or a simple multiple of the ratio of the weights of A and which directly combine with each other. Gay Lussac s law of gaseous volumes : When gases react together, they always do so in volumes which bear a simple ratio to one another and to the volumes of the products, if gaseous, all measurements are made under the same conditions of temperature and pressure. Avogadro s hypothesis : Atom : Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules. It is the smallest particle of an element that takes part in a chemical reaction. It may or may not be capable of free existence. Atomic mass unit (a.m.u) : It is equal to 1 1 th of the mass of an atom of C-1. It is equal to g. Atomic mass : It is the average relative mass of its atoms as compared with an atom of carbon 1 isotope taken as 1. Calculation of average atomic mass : If an element exists in two isotopes having atomic masses a and b in the ratio m: n, then average atomic mass = m a n b. m n Gram atomic mass (G.A.M.): The atomic mass of an element expressed in grams is called gram atomic mass. This amount of the element is called one gram atom. This amount of the element can also be called one mole atom.

2 Method of determining atomic weight (i) (ii) y application of the relation At. wt. = Eq. wt. Valency Knowing the exact eq. wt. and approximate atomic weight, valency can be calculated (which is a whole number). Knowing exact eq. wt. and valency, exact atomic weight can be calculated. Dulong and Petit s method: According to Dulong and Petit s law, for solid elements (except e,, C and Si) At. wt. Specific heat = 6.4 (approx.) Approx at. wt. = 6.4 Sp. heat = Approx. At. wt. Eq. wt. = approx. Valency. The approx. valency is converted to the nearest whole number which is called exact valence. Eq. wt. exact valency = exact Atomic Weight. (iii) Specific heat/molar heat capacity method: For gases, Cp is 1.66 for monoatomic, 1.40 for C diatomic and 1.30 for triatomic gases. Thus having known the atomicity of the gas from the value Mol. wt of, Atomic wt. of the gaseous element = Atomicitẏ v Molecular wt. = Vapour density. (iv) Method: This method can be used for those elements whose chlorides are volatile so that their vapour densities can be determined. Then Mol. wt. of the chloride = V.D. If x is the valency of the element (M), then the formula of its chloride will be MCl x. Hence Molecular weight of the Chloride MCl x = At. wt. of M + x 35.5 = Eq. wt. of M Valency of M + x 35.5 = E x + x 35.5 = x (E ) x (E ) = V.D. or x = E V.D Knowing the eq. wt. E of the element, the valency x can be calculated. Then Atomic weight = eq. wt. Valency. (v) Law of Isomorphism: Compounds having similar molecular formulae and identical crystal structure are called isomorphs. The method is based upon the fact that elements in isomorphous compounds have same valencies e.g., (a) K SO 4, K CrO 4 and K SeO 4 are isomorphous. Hence valency of S, Cr and Se =

3 (b) (c) ZnSO 4.7H O, FeSO 4.7H O, MgSO 4.7H O are isomorphous. Hence valency of Zn, Fe and Mg =. Alums, M SO 4.M (SO 4 ) 3.4H O in which M is monovalent and M is trivalent are isomorphous. Knowing the valency, Atomic wt. = Eq. wt. Valency Molecules : It is the smallest particle of an element or a compound that is capable of free existence. Molecular mass : Molecular mass of a substance is the average relative mass of its molecules as compared with an atom of C 1 isotope taken as 1. Gram molecular mass or molar mass (G.M.M.): The molecular mass of a substance expressed in gram is called gram molecular mass or molar mass. This amount of the substance is called one gram molecule. This amount of the substance is called one mole of the substance. Methods of determining molecular weight (i) Gram Molecular Volume (G.M.V.) method,.4 litres of every gas or vapour at STP weigh equal to molecular weight expressed in grams. This is the principle of Victor Meyer method used for volatile liquids. (ii) Vapour density method Molecular weight = Vapour Density where Vapour Density = Wt. of Wt. of same volume of certain volume of the vapour under same conditions of temp. and H pressure (iii) Diffusion method: According to Graham s law of diffusion, rates of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to the square root of their densities or molecular weights i.e., r r 1 d d 1 d d 1 M M 1 Knowing the molecular weight of one gas, that of the other can be calculated. (iv) Colligative property method: Discussed in the unit of Solutions. Moles : The collection of molecules of an element or ions or compounds constitute one mole of that element, ion or compound. This number, is known as Avogadro s number. This is called mole-particle relationship. Mole-Particle Relationship : e.g. (i) one mole of Na contains atoms of Na. (ii) one mole of oxygen i.e. one mole of O contains molecules of oxygen

4 (iii) one mole of CCl 4 carbon tetrachloride contains molecule of CCl 4 (carbon tetrachloride) Mole-Weight Relationship : (i) one mole of an element weighs equal to gram atomic weight of the element e.g. one mole of Na weighs 3 gm (Gram atomic wt. of Na = 3) (ii) one mole of Mg weighs 4 gms (GAM of Mg = 4) (iii) one mole of substance (molecular state) weighs equal to the gram molecular weight of the substance. e.g. one mole of O weighs 3 gms (GMM of O = 3) one mole of SO weighs 64 gms. Mole-Volume Relationship : (i) one mole of every gas at STP occupies.4 litres of volume. e.g. one mole of CO at STP occupies.4 litres of volume. (ii) one mole of SO at STP occupies.4 litres of volume. GRAM-ATOMS : One gram-atom of an element means collection of atoms. This concept applies only to the elements which exists in polyatomic states (e.g. O, Cl, S, P etc.). It is meaningless for metals and compounds. * The number of atoms in w gms of an element whose atomic mass is A is : w gm atom = mole atom A * The number of atoms is given by: Avogadro s number : w No. of atoms = N0 A It is the number of atoms present in one gram atom of an element or the number of molecules present in one gram molecule of the substance. In general, it is the number of particles present in one mol of the substance. Its value is Equivalent weight : The equivalent weight of a substance is the number of parts by weight of the substance that combine with or displace directly or indirectly parts by weight of hydrogen or 8 parts by weight of oxygen or 35.5 parts by weight of chlorine. Mol. wt. of the acid Eq. wt. of an acid = asicity Mol. wt. of the base Eq. wt. of a base = acidity Gram equivalent weight : The equivalent weight of a substance expressed in grams is called gram eq. wt. or one gram equivalent. Now we can define gram-equivalent (gm eq) in g gms of a substance whose equivalent weight is E as follows: No. of e q = E g - 4 -

5 * The number of milliequivalents (meq) in g gms is given by: m e q = E g 1000 Methods of determining equivalent weight (i) Hydrogen displacement method: This method is used for metals which react with an acid to evolve hydrogen gas. Eq. wt. of the metal is the weight of the metal which displace g of H or 1100 c.c. of H at STP. Wt. of metal Eq. wt. of metal = Wt. of H displaced Wt. of metal or = 1100 Vol. of H in ml displaced at STP (ii) Oxide formation or reduction of the oxide method: In this method a known weight of the metal is converted into its oxide directly or indirectly. Knowing the weight of the metal oxide formed, the weight of oxygen combined can be calculated. Alternatively, a known weight of the metal oxide may be reduced to metal whose weight is determined. Wt. of metal Eq. wt. of metal = 8 Wt. of oxygen combined Wt. of metal or = 5600 Vol. of O displaced / combined in ml at STP Eq. wt. of metal = weight of metal Wt. of chlorine combined 35.5 Wt. of metal or 1100 Vol. of Cl combined in ml at STP (iii) Double decomposition method: For a reaction of the type A + CD AD + C (e.q. AgNO 3 + NaCl AgCl + NaNO 3 ) Weight of A taken Weight of AD formed = Eq. wt. of Eq. wt. of A AD Eq. wt. of A Eq. = Eq. wt. of A wt. of Eq. wt. of D Knowing the equivalent weights of any two radicals out of A, and D, that of the third can be calculated. (iv) Electrolytic method (based on Faraday s laws of electrolysis) 1 Faraday (96500 coulombs) deposit one gram equivalent of the substance. Electrochemical equivalent of a substance is the weight of the substance deposited by one coulomb

6 (v) (vi) When the same quantity of electricity flows through solutions of different electrolytes, Wt. of Wt. of X deposited Y deposited = Eq. wt. of X Eq. wt. of Y Neutralization method: This is based on the fact that acids and bases react in equivalent amounts. Hence equivalent weight of an acid is the weight of the acid which is neutralized by 1000 cc of 1N base solution (which contains 1g eq. of the base) Likewise equivalent weight of a base can be determined. Silver salt method: (for organic acids only) The organic acid is converted into its silver salt (RCOOAg), a known weight of which is ignited to give residue of Ag which is weighed. Then Eq. wt. of Eq. wt. of RCOOAg Ag (108) = Wt. of Wt. of silver salt silver Eq. wt. of acid (RCOOH) = Eq. wt. of RCOOAg 108. (vii) Conversion method: When one compound of a metal is converted into another compound of the same metal (e.g. metal carbonate metal oxide), then Weight of compound I Weight of compound II = Eq. wt. of Eq. wt. of metal Eq. wt. of anion of compound I metal Eq. wt. of anion of compound II Relationship between eq. wt. atomic wt. and valency of an element: Eq. wt. = Atomic weight Valency Equivalent weight of oxidizing/reducing agent = Expressing concentration of solutions Molecular weight of the substance No. of electrons gained/lost by one molecule Solution is a homogenous mixture of two or more components in which Intermixing particles are of atomic or molecular dimensions. A solution consists of a dissolved substance known as solute and the substance in which the solute is dissolved is known as solvent. The concentration of a solution means the quantity of solute dissolved per unit volume of solution, or per unit quantity of solvent. concentration of solution = quantity of solute quantity of solvent or solution Note: While discussing various methods for expressing concentration, we have taken solute as dissolved in solvent A and W A as grams of solute and W as grams of solvent. Mass fraction is the fractional part of a component that is contributed by it to the total mass of solution. mass fraction of = w w w A mass fraction of A = w A w w A Mole fraction is the fractional part of the moles that is contributed by each component to the total number of moles that prepare the solution. In containing n A moles of solvent and n moles of solute: - 6 -

7 mole fraction of = x = mole fraction of A = x A = n n n n A A n A n Molality (m) is expressed as number of moles of solute dissolved in 1000 gms (1 Kg) of solvent. It is denoted by m. m n w A 1000 Molarity (m) is expressed as moles of solute contained in one litre of solution or it is also taken as millimoles of solute in 1000 cc (ml) of solution. It is denoted by M. molarity (M) = moles of solute litres of solution = millimoles of solute millilitres of solution n M V = lt W / W V (lit) is expressed as the number of gram equivalents (gm eq) of solute contained in one litre of solution or it can also be taken as number of milliequivalents (meq) in 1000 cc (ml) of solution. It is denoted by N. normality of solution (N) = N = gm eq of solute litres of solution gm eq V (lit) = meq of solute millilitres of solution Note : The following results should be remembered and used directly. Molarity (M) and Normality (N) are temperature dependent but molality (m) and mole fraction are Temperature independent units. (1) moles = M V lt = w M w milimoles = M V cc = 1000 M () gm-eq = N V lt = w E w meq = N V cc = 1000 E 10 xd (3) Molarity (M) = M d : density of solution in g/cc (4) N = x M (5) 1000 x m = M x A A 0 Normality (N) = 10 xd E x :% age strength of solution (by weight) x : acidity for base or basicity for acid or electron transfer / mole for O.A. and R.A./n-factor

8 Strength of a solution is generally expressed as grams of solute in one litre of solution. strength = Diluting a solution: w litres of solution strength = Normality Eq. wt. = Molarity Mole wt. Whenever a given solution of known concentration i.e. normality & molarity (known as standard solution) is diluted by adding more of solvent, the number of millimoles (or millequivalents) of solutes remain unchanged. The concentration of solution however changes. In such cases M 1 V 1 = M V (M 1 & V 1 a re molarity and volume of original solution) (M & V are molarity and volume of diluted solution) STOICHIOMETRIC CALCULATIONS: (Quantitative analysis of a balance chemical equation) Consider a balanced chemical reaction of equation: ma + n pc + qd A and are reactants; C and D are products; m, n, p, q are the stoichiometric coefficients. The above balanced reaction is analysed as: m moles of A react with n moles of to produce p moles of C and q moles of D. This can be represented as: m moles of A n moles of p moles of C q moles of D It is also observed that when a substance takes part in a chemical reaction then the amount of the substance taking part is proportional to its equivalent weight i.e. 1 equivalent of every substance combines (displaces, double displaces, oxidises, reduces, neutralizes) are equivalent of another substance. Hence in a chemical reaction of the type aa + b mc + nd 1 eq. of A combines with 1 eq. of to form 1 eq of C and 1 eq of D or n eq of A n eq of n eq of C n eq of D. In general, there are two types of stoichiometric reactions (which are most common): Neutralisation Reactions The analysis of two types of reactions is generally carried out in the form of mass of reactants (or products) taking part in a given reaction (gravimetric analysis) or in terms of concentrations of reactants (or products) taking part in a given reaction (volumetric analysis)

9 Neutralisation: A reaction in which an acid (or a base) completely reacts with a base (or an acid) to form salt and water. If HA be the acid, OH be the base and A be the salt, then neutralisation reaction can be represented as follows: HA + OH A + H O As we known that an acid may be monobasic (HCl, HNO 3, etc.), dibasic (H SO 4, H C O 4 etc.) or tribasic (H 3 PO 4 etc.) and similarly a base may be monoacidic (NaOH, NaHCO 3 etc.), diacidic (Ca(OH), Na CO 3 etc.) or triacidic (Al(OH) 3 etc.), so it is better to define the neutralisation reaction in the following manner. A reaction in which 1 gram equivalent (or 1 meq of an acid (or a base) completely reacts with 1 gram equivalent (or 1 meq) of a base (or an acid) to form 1 gram equivalent (or 1 meq) of corresponding salt

10 SOLVED EXAMPLES Example 1 :Fe (SO 4 ) 3 is used in water and sewage treatment to aid the removal of suspended impurities. Calculate the mass percentage of iron, suphur and oxygen in this compound. Solution : Step-I: Molecular weight of Fe (SO 4 ) 3 = (56 ) + (3 3) + (16 1) = 400 Number of parts by weight of Fe Step-II: % of Fe = 100 Molecular weight of Fe (SO ) = = 8% % of S = Numberof parts by weight of S 100 Molecular weight of Fe (SO ) 4 3 = = 4% 400 % of O = Number of parts by weight of O 100 Molecular weight of Fe (SO ) 4 3 = = 48% 400 Example :An inorganic salt gave the following percentage composition: Na = 9.11%, S = 40.51% and Solution : O = 30.38%. Calculate the empirical formula of the salt. Calculation of empirical formula. Element Symbol % of element At. Mass of Relative no. of atoms Simplest atomic ratio Simplest whole element = percentage number atomic mass atomic ratio Sodium Na Sulphur S Oxygen O Thus, the empirical formula is Na S O 3. Example 3: Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO 3 ). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O ). How many moles and how many grams of KClO 3 are required to produce.4 mole O? (M KClO = 1.5g/ mole) 3 Solution : Decomposition of KClO 3 takes place as KClO 3 (s) KCl(s) + 3O (g)

11 mole KClO 3 gives 3 mole of O 3 mole O is formed by mole of KClO 3..4 mole O will be formed by.4 3 mole KClO 3 = 1.6 mole KClO 3. Mass of KClO 3 = Number of moles Molar mass = = 196 g Example 4 :1 litre mixture of CO and CO is taken. This is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. The volumes are measured under the same conditions. Find the composition of mixture by volume. Solution : Let there be x ml of CO in the mixture. Hence there will be (1000 x) ml CO. The reaction of CO with red hot charcoal may be given as, CO (g) + C(s) CO(g) 1000 x 000 x Total volume of the gas becomes = x + (1000 x) Volume of CO = 400 ml and volume of CO = 600 ml x x = 1600 x = 400 ml Example 5 :A solution of oxalic acid C H O 4. H O is prepared by dissolving 0.63 g of the acid in 50 cm 3 Solution : of the solution. Calculate molarity of the solution. Molar mass of oxalic acid = 16 g/mol. 50 cm 3 or 50 L = 0.5 L of the solution contains 0.63 g oxalic acid Molarity of the solution = = 0.0 M

12 OJECTIVE QUESTIONS 1. For the reaction A + C, 5 moles of A and 8 moles of will produce (a) 5 moles of C (b) 4 moles of C (c) 8 moles of C (d) 13 moles of C. An hydride of nitrogen decomposes to give nitrogen and hydrogen. It was formed that one volume of the hydride gave one volume of N and volumes of H at STP. The hydride of nitrogen is (a) NH 3 (b) N H 6 (c) NH (d) N H volumes of a hydrocarbon on complete combustion consumed 10 volumes of oxygen giving 5 volumes of CO at STP. The hydrocarbon is (a) C H 6 (b) C H 4 (c) CH 4 (d) C H 4 4. The volume of oxygen used when x gms of Zn is converted to ZnO is (a) x 5.6 litres 65 (b) x litres (c) 4 x litres (d) None of these 5. 1 g of Mg will react completely with an acid to give (a) 1 mole of O (b) 1 mole of H (c) 1 mole of H (d) moles fo H 6. The volume of water which should be added to 300 ml of 0.5 M NaOH solution so as to get a solution of 0. M is: (a) 550 ml (b) 350 ml (c) 750 ml (d) 450 ml 7. The mole fraction of a solution containing 3.0 gms of urea per 50 gms of water would be (a) (b) (c) (d) None of these 8. The molality of a solution of conc. HCl containing 36.5% by weight of HCl would be (a) (b) (c) (d) A mixture of N and H to react in a closed container to form NH 3. The reaction ceases before either reactant has been totally consumed. At this stage,.0 moles each of N, H and NH 3 are present. The moles of N and H present originally were respectively, (a) 4 and 4 moles (b) 3 and 5 moles (c) 3 and 4 moles (d) 4 and 5 moles 10. The number of atoms present in 16 g of oxygen is (a) (b) (c) (d)

13 litres of oxygen at NTP is equivalent to (a) 1 mole (b) 1 mole (c) 1 4 mole (d) 1 8 mole 1. How many grams are contained in 1 gram-atom of Na? (a) 13 g (b) 3 g (c) 1 g (d) 1 3 g 13. How many moles of electrons weigh one kilogram? (a) (b) (c) (d) The haemoglobin form the red corpuscles of most mammals contains app. 0.33% of iron by weight. The physical measurement indicates the molecular weight of haemoglobin The number of iron atoms in each molecule of haemoglobin is (At. wt. of Fe = 56) (a) (b) 3 (c) 4 (d) None of these 15. If 0.5 mol of acl is mixed with 0. mol of Na 3 PO 4, the maximum number of mole of a 3 (PO 4 ) that can be formed is (a) 0.7 (b) 0.5 (c) 0.30 (d) Number of HCl molecules present in 10 ml of 0.1 N HCl solution is (a) (b) (c) (d) A gas of mixture of 3.0 litres of propane and butane on complete combustion at 5ºC produced 10 litres of CO. Find out the composition of the gas mixture. (a) 70%, 30% respectively (b) 66.66%, 33.33% respectively (c) 30%, 70% respectively (d) None of these 18. No. of electrons in 1.8 ml of H O(l) is (a) (b) (c) (d) One mole of a mixture of CO and CO required exactly 0 g of NaOH in solution for complete conversion of all the CO into Na CO 3. How much NaOH would it require for conversion into Na CO 3 if the mixture (one mole) is completely oxidised to CO : (a) 60 g (b) 80 g (c) 40 g (d) 0 g 0. The number of H + ions present in 100 ml of M H SO 4 solution is (a) (b) (c) (d)

14 1. The amount of wet NaOH containing 0% water required to neutralise 6 litre of 0.5 M H SO 4 solution is (a) 3 kg (b) 1.5 kg (c) 0.3 kg (d) 0.15 kg. 1 g of an acid (mol. wt. 146) is completely neutralised by g KOH. The number of neutralisable protons in acid are (a) 1 (b) 3 (c) (d) 4 3. The volume ratio of 6 N and N HCl required to prepare 100 ml of 5 N HCl is (a) 3 : 1 (b) 1 : 3 (c) 4 : 1 (d) 1 : ml of pure water at 4 C is saturated with NH 3 producing a solution of 30% by mass of NH 3. The total weight of solution after saturation becomes: (a) 105 g (b) 130 g (c) 150 g (d) 160 g 5. The normality of 0.3 M phosphorous acid: (a) 0.1 (b) 0.9 (c) 0.3 (d) How many gram of dibasic acid (mol. wt. 00) should be present in 100 ml of the aqueous solution to give 0.1 N solution: (a) 1 g (b) 1.5 g (c) 0.5 g (d) 0 g 7. Specific volume of a cylindrical virus particle is ml/g, having the radius and length 7Å and 10Å respectively. What is the molecular weight of virus: (a) 15.4 kg/mol (c) kg /mol (b) kg /mol (d) kg /mol 8..5 litre of 1 M NaOH solution is mixed with another 3 litre solution of 0.5 M NaOH. The molarity of resultant solution is (a) 1.0 M (b) 0.84 M (c) 0.73 M (d) 0.56 M 9. Select the incorrect statement for exact neutralisation of 1 mole of a(oh) by (a) 1.5 mole H 3 PO 3 (b) 1 mole of H SO 4 (c) /3 mole of H 3 PO 4 (d) mole of H 3 PO 30. When the same amount of Zn is treated separately with excess of H SO 4 and excess of NaOH, the ratio of volumes of H evolved is (a) 1 : 1 (b) 1 : (c) : 1 (d) 9 :

15 MORE THAN ONE CORRECT ANSWERS TYPE mole of a mixture of CO and CO requires exactly 1 litre solution of 1 M NaOH for complete neutralisation. If CO present in mixture is now converted to CO and again the mixture is treated with NaOH, then after this conversion: (a) moles of CO present initially in mixture =1 (b) litre NaOH solution of 1 M is more required for neutralisation (c) litre solution of 1/ M NaOH is required more fore neutralisation (d) 56 g KOH in aqueous solution is required more for neutralisation 3. g of oleum is diluted with water. The solution was then neutralised by 43.5 ml of 0.1 N NaOH. Select the correct statements: (a) % of oleum is (b) % of free SO 3 is 6.5 in oleum (c) Equivalent of H SO 4 are 0.03 (d) Equivalent of SO 3 = Which one is not correct about VO + Fe O 3 FeO + V O 5? (a) moles of VO reacts completely with 5 mole of Fe O 3 (b) 1 mole of VO reacts completely with 1.5 mole of Fe O 3 (c) Eq. weight of V O 5 = M/6 and of Fe O 3 is M/ (d) Eq. weight of VO = M/3 and of FeO is M/ mole of H 3 PO 3 reacts with NaOH in solution. Select the correct statements: (a) 1 mole of NaOH will replace N H + ion from H 3 PO 3 (b) moles of NaOH will replace N H + ion from H 3 PO 3 (c) 3 moles of NaOH will replace 3N H + ion from H 3 PO 3 (d) on complete neutralisation of H 3 PO 3, the equivalent weight of H 3 PO 3 = ml of 0.8 M NaOH are used to neutralised 100 ml solution obtained by passing.70 g SO Cl in water. Select the correct statement: (a) The solution of SO Cl has 0. M H SO 4 and 0.4 M HCl (b) The volume ratio of NaOH used for H SO 4 and HCl is 1 : (c) The volume ratio of NaOH used for H SO 4 and HCl is 1 : 1 (d) Molarity of SO Cl solution is 0.1 M 36. Which one are correct about the solution that contains 3.4 ppm Al (SO 4 ) 3 and 1.4 ppm Na SO 4? (a) [Al 3+ ] = [Na + ] (b) [SO 4 ] = [Na + ] = [Al 3+ ] (c) [SO 4 ] = [Na + ] + [Al 3+ ] (d) [SO 4 ] = [Na + ]

16 37. 1 U is equal to (a) MeV (b) g (c) 1/1 mass of C 1 (d) 1 dalton 38. 1/N A is equal to (a) kg (b) 1 U (c) 1 dalton (d) Logschmidt number 39. H 3 O 3 is (a) Monobasic acid (c) Electron pair acceptor (b) Lewis acid (d) Na 3 O 3 exist as ionic compound 40. Which of the following are primary standard solution: (a) Oxalic acid (b) NaOH (c) orax (d) Na CO 3. 10H O

17 MISCELLANEOUS ASSIGNMENT Comprehension-I Normality is number of gram equivalents dissolved per litre of solution. It changes with change in temperature. In case of monobasic acid, normality and molarity are equal but in case of dibasic acid, normality is twice the molarity. In neutralization and redox reactions, number of milliequivalents of reactants as well as products are always equal 1. On heating a litre of a N/ HCl solution,.750 g of HCl is lost and the volume of solution becomes 750 ml. The normality of resulting solution will be (a) 0.58 (b) 0.75 (c) (d) 5.7. The volume of 0.1 M Ca(OH) required to neutralize 10 ml of 0.1 N HCl will be (a) 10 ml (b) 0 ml (c) 5 ml (d) 40 ml 3. Molarity of 0.5 N Na CO 3 is (a) 0.5 (b) 1.0 (c) 0.5 (d) N KOH solution in water contains 30% by weight of KOH. The density of solution will be (a) 1.88 (b).88 (c) (d) 1.88 Comprehension-II The density of 3 M solution of Na S O 3 is 1.5 g ml 1 5. The % by weight of Na S O 3 is (a) 36.4 (b) 37.9 (c) 40.4 (d) Mole fraction of Na S O 3 is (a) (b) 0.05 (c) (d) Molalities of Na + and S O 3 ions are respectively (a) 7.73, (b) 3.866, 7.73 (c) 3.73, (d) 7.866, Column I contains the compounds of Ammonia which will react with 4 gm of KOH to evolve NH 3. Column II contains the volume of Ammonia evolved in the reaction. Match Column I with Column II. Column I Column II A. 149 gm of (NH 4 ) 3 PO 4 (p) gm of NH 4 ClO (q) 89.6 C. 166 gm of NH 4 H PO (r).4 D. 58 gm of (NH 4 ) SO 3 (s) 67.

18 9. X, Y are two elements which has equivalent weight 40 and 80 respectively. List I contains the corresponding compounds of X and Y and List II contains the ratio of their equivalent weight of X and Y of the corresponding compounds. Column I Column II A. Sulphite (p) Nitrate (q) 0.75 C. Carbonate (r) 0.7 D. romide (s) 0.67 INTEGER TYPE QUESTIONS g of a polybasic acid (mol. wt. 98) requires 30 ml of 0.5 M a(oh) for complete neutralisation. The basicity of acid is 11. A solution of H O has normality N/1.7. Its % strength is 1. Number of water molecules attached on Cu + ion in CuSO 4. 5H O is 13. n factor for potash alum is 14. A gaseous alkane C n H n + is exploded with oxygen. The volume of oxygen for complete combustion of alkane to CO formed is in the ratio 7 : 4. The value of n is. 15. Two acid solution A and are titrated separately each with 5 ml of 1 N Na CO 3 solution. The volume of each acid used for titration is 10 ml and 40 ml respectively for the complete neutralisation. The volume ratio of V and V A which is mixed to prepare one ml 1 N solution is. 16. Number of equivalents of oxygen in 1 M K Cr O 7 acting as oxidant in presence of acid are. 17. An inorganic compound ZnCr O x, 9.81 g Zn, atoms of Cr and 0.6 g atom of O. The volume of x is. 18. The equilibrium molarity of OH is 0.08 M in a solution of 0.1 M (OH). The % of dissociation of Ca(OH) is. 19. The normality of 0.5 M H 3 PO 3 solution is

19 PREVIOUS YEAR QUESTIONS IIT-JEE/JEE-ADVANCE QUESTIONS 1. How many moles of electron weigh one kilogram? (a) (b) (c) (d) The total number of electrons present in 18 ml of water (density of water is 1 g ml 1 ) is (a) (b) (c) (d) If 10 1 molecules are removed from 00 mg of CO, then the number of moles of CO left are (a) (b) (c) (d) The number of gram molecules of oxygen in CO molecules is (a) 10 gm molecules (b) 5 gm molecules (c) 1 gm molecules (d) 0.5 gm molecules 5. At 100 C and 1 atm, if the density of liquid water is 1.0 g cm 3 and that of water vapour is g cm 3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (a) 6 cm 3 (b) 60 cm 3 (c) 0.6 cm 3 (d) 0.06 cm 3 6. The triad of nuclei which is isotonic is (a) 6 C14, 7 N 15, 9 F 17 (b) 6 C1, 7 N 14, 9 F 19 (c) 6 C14, 7 N 14, 9 F 17 (d) 6 C14, 7 N 14, 9 F % (w/w) HCl stock solution has a density of 1.5 g ml 1. The molecular weight of HCl is 36.5 g mol 1. The volume (ml) of stock solution required to prepare a 00 ml solution of 0.4 M HCl is 8. If the value of Avogadro number is mol 1 and the value of oltzmann constant is J K 1, then the number of significant digits in the calculated value of the universal gas constant is 9. A compound H X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g ml 1. Assuming no change in volume upon dissolution, the molality of a 3. molar solution is 1. Mole concept is not applicable to (a) volume of gases as NTP (b) volume of liquids at NTP (c) atoms & molecules (d) ions & particles. The dot at the end of a sentence has a mass of one microgram. Assuming the black stuff is carbon, calculate the number of carbon atoms which make such a dot (a) 10 3 (b) (c) (d) A mixture of CH 4 N and O is inclosed in a vessel of one litre capacity at 0 C. The ratio the partial

20 pressure of gases is 1 : 4 :. The total pressure of the gaseous mixture is 660 mm. The number of molecules of oxygen present in vessel is (a) (b) (c).4 10 (d) An X molal solution of a compound in benzene has mole fraction of solute equal to 0.. The value of X is (a) 14 (b) 3. (c) 1.4 (d) 5. X gm of an element gave Y gm of oxide. Eq. wt. of the element is X (a) 8 Y X X (b) (Y X) 8 (c) 8 Y X X Y (d) 8 X 6. If 9.8 gm of hexane burns completely in oxygen, how many moles of CO is produced? (a) 6 (b) 0.6 (c) 0.9 (d) If m 1 gm of metal A displaces m gm of another metal from the salt solution and if their equivalent weights are E 1 and E respectively, then the equivalent weight can be expressed by (a) E 1 = m 1 E (b) E 1 = m m E (c) E 1 = m 1 m1 m E m1 (d) E 1 = E m 8. What volume of CO will be liberated at NTP if 1. gm of carbon is burnt in excess of oxygen? (a) 11. litres (b).4 litres (c).4 litres (d) 1.1 litres 9. One mole of CO contains (a) atoms of C (b) atoms of O (c) molecules of CO (d) 3 gm atom of CO 10. The Avogadro Number or a mole represents (a) ions (b) atoms (c) molecules (d) entities 11. What is the weight of one molecule of a monatomic element X whose atomic weight is 36? (a) gm (b) gm (c) gm (d) gm 1. Air contains 0% O for the reaction C H 5 NH + O CO + NO + H O. How much volume of air will be required for 1 mol of C H 5 NH (a) L (b) 1064 L (c) 1.8 L (d) 18 L 13. Oxide of Mn contains 70% of Mn formula of this oxide will be - 0 -

21 (a) Mn O 3 (b) MnO (c) Mn 3 O (d) Mn O 14. Under certain thermodynamic conditions atomic wt of oxygen is 16.7, it contains two isotopes 16 8 O and 17 8 O. Percentage 17 8 O isotopes is (a) 50 (b) 60 (c) 70 (d) % by weight solution will contain how much mass of the solute in 1 litre solution, density of the solution is 1. g/ml (a) 480 g (b) 48 g (c) 38 g (d) 380 g AIEEE/JEE-MAINS QUESTIONS 1. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 50 ml of 1. M second solution. What is the molarity of the final mixture? (a) 1.0 M (b) 1.50 M (c) M (d).70 M. If we consider that 1/6, in place of 1/1, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of substance will (a) decrease twice (b) increase two fold (c) remain unchanged (d) be a function of the molecular mass of the substance molecules of urea are present in 100 ml of its solution. The concentration of urea solution is (a) 0.0 M (b) 0.01 M (c) M (d) 0.1 M (Avogadro constant, N A = mol 1 ) 4. To neutralise completely 0 ml of 0.1 M aqueous solution of phosphorous acid (H 3 PO 3 ), the volume of 0.1 M aqueous KOH solution required is (a) 40 ml (b) 0 ml (c) 10 ml (d) 60 ml 5. 5 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid volume gave a of 35 ml. The molarity of barium hydroxide solution was (a) 0.14 (b) 0.8 (c) 0.35 (d) For an ideal gas, number of moles per litre in terms of its pressure P, gas constant R and temperature T is (a) PT/R (b) PRT (c) P/RT (d) RT/P 7. In a compound C, H and N atoms are present in 9 : 1 : 35 by weight. Molecular weight of compound - 1 -

22 is 108. Molecular formula of compound is (a) C H 6 N (b) C 3 H 4 N (c) C 6 H 8 N (d) C 9 H 1 N 3 8. With increase of temperature, which of these changes? (a) Molality (b) Weight fraction of solute (c) Fraction of solute present in water (d) Mole fraction 9. Number of atoms in gram Fe (at wt. of Fe = g mol 1 ) is (a) twice that in 60 g carbon (b) (c) half that in 8 g He (d) How many moles of magnesium phosphate, Mg 3 (PO 4 ) will contain 0.5 mole of oxygen atoms? (a).5 10 (b) 0.0 (c) (d) In the reaction, Al (s) + 6HCl (aq) Al 3+ + (aq) 6Cl + 3H, (aq) (g) (a) 67. L H (g) at STP is produced for every mole Al that reacts (b) 11. L H (g) at STP is produced for every mole HCl (aq) consumed (c) 6 L HCl (aq) is consumed for every 3L H (g) is produced (d) 33.6 L H (g) is produced regardless of temperature and pressure for every mole Al that reacts 1. The density (in g ML 1 ) of a 3.60 M sulphuric acid solution that is 9% H SO 4 (Molar mass = 98 g mol 1 ) by mass will be (a) 1. (b) 1.45 (c) 1.64 (d) A gaseous hydrocarbon gives upon combustion 0.7 g of water and 3.08 g, of CO. The empirical formula of the hydrocarbon is (a) C 6 H 5 (b) C 7 H 8 (c) C H 4 (d) C 3 H The molarity of a solution obtained by mixing 750 ml of 0.5 (M)HCl with 50 ml of (M) HCl will be: (a) 1.75 M (b) M (c) M (d) 1.00 MN - -

23 SUJECTIVE PROLEMS 1. 3 g of a salt of molecular weight 30 is dissolved in 50 g of water. Calculate the molality of the solution.. How much acl.h O and pure water are to be mixed to prepare 50 g of 1.0% (by wt.) acl solution. 3. Calculate the total number of atoms in 0.5 mole of K Cr O Find the ratio of the number of molecules contained in 1 g of NH 3 and 1 g of N. 5. Calculate the number of oxygen atoms in 0. mole of Na CO 3. 10H O. 6. A precipitate of AgCl and Agr weighs g. On heating in a current of chlorine, the Agr is converted to AgCl and the mixture loses g in weight. Find the percentage of Cl in original mixture. 7. Three atoms of magnesium combine with atoms of nitrogen. What will be the weight of magnesium which combines with 1.86 g of nitrogen? ml of a mixture of O 3 and O weighs 1 g at NTP. Calculate the volume of ozone in the mixture. 9. A solid mixture 5 g consists of lead nitrate and sodium nitrate was heated below 600ºC until weight of residue was constant. If the loss in weight is 8%, find the amount of lead nitrate and sodium nitrate in mixture g of a mixture of K CO 3 and Li CO 3 required 30 ml of 0.5 N HCl solution for neutralization. What is percentage composition of mixture? ml of 8 N HNO 3, 4.8 ml of 5 N HCl and a certain volume of 17 M H SO 4 are mixed together and made upto litre. 30 ml of this acid mixture exactly neutralizes 4.9 ml of Na CO 3 solution containing 1 g Na CO 3.10H O in 100 ml of water. Calculate the amount of sulphate ions in g present in solution. 1. How many litres of liquid CCl 4 (d = 1.5 g/cc.) must be measured out to contain CCl 4 molecules? 13. A mixture of ethane (C H 6 ) and ethene (C H 4 ) occupies 40 litre at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O to produce CO and H O. Assuming ideal gas behaviour, calculate the mole fractions of C H 4 and C H 6 in the mixture. 14. An evacuated glass vessel weighs 50 gm. when empty, 148.0g when completely filled with liquid of density 0.98g ml 1 and 50.5g when filled with an ideal gas at 760mm at 300 K. Determine the molecular weight of the gas.

24 15. Calculate the molarity of pure water using its density to be 1000 kg m One gm. of charcoal absorbs 100 ml. 0.5 M CH 3 COOH to form a monolayer, and thereby the molarity of CH 3 COOH reduces to Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = m /gm. 17. Calculate the amount of calcium oxide required when it reacts with 85gm. of P 4 O 10. 6CaO + P 4 O 10 Ca 3 (PO 4 ) 18. 0% surface sites have adsorbed N. On heating N gas evolved from sites and were collected at atm and 98K in a container of volume is.46 cm 3. Density of surface sites is /cm and surface area is 1000 cm, find out the no. of surface sites occupied per molecule of N. 19. For a hypothetical chemical reaction represented by : 3A(g) C(g) D(g), the following informations are known. Information (i) At t = 0, only 1 mole of A is present and the gas has VD. = 60 (ii) At t = 30 min., the gaseous mixture consist of all three gases and has a vapour density = 75. (iii) Molecular mass of C = 00 Calculate (a) Molecular weight of A and D. (b) Moles of each specie at t = 30min. 0. A mixture of Ferric oxide (Fe O 3 ) and Al is used as a solid rocket fuel which reacts to give Al O 3 and Fe. No other reactants and products are involved. On complete reaction of 1 mole of Fe O 3, 00 units of energy is released. (i) Write a balance reaction representing the above change. (ii) What should be the ratio of masses Fe O 3 and Al taken so that maximum energy per unit mass of fuel is released. (iii) What would be energy released if 16 kg. of Fe O 3 reacts with.7 kg of Al

25 ANSWERS Objectives 1. (b). (d) 3. (c) 4. (a) 5. (b) 6. (d) 7. (a) 8. (c) 9. (b) 10. (d) 11. (c) 1. (b) 13. (d) 14. (c) 15. (d) 16. (d) 17. (b) 18. (a) 19. (b) 0. (d) 1. (c). (c) 3. (a) 4. (c) 5. (d) 6. (a) 7. (a) 8. (a) 9. (a) 30. (a) 31. (c,d) 3. (a,b,c) 33. (b,c,d) 34. (a,b,d) 35. (a,c) 36. (a,c) 37. (a,b,c,d) 38. (a,b,c) 39. (a,b,c,d) 40. (a,c,d) Miscellaneous Assignment 1. (a). (c) 3. (a) 4. (a) 5. (b) 6. (c) 7. (a) 8. A-(s); -(q); C-(p); D-(r) 9. A-(s); -(r); C-(p); D-(q) 10. (3) 11. (1) 1. (4) 13. (8) 14. () 15. (4) 16. (6) 17. (4) 18. (4) 19. (3) Priviouse Year IIT-JEE/JEE-ADVANCE QUESTIONS 1. (d). (c) 3. (a) 4. (b) 5. (c) 6. (a) 7. (8) 8. (4) 9. (8) DCE QUESTIONS 1. (b). (b) 3. (a) 4. (b) 5. (c) 6. (b) 7. (a) 8. (c) 9. (a) 10. (d) 11. (a) 1. (b) 13. (a) 14. (c) 15. (a) MAINS QUESTIONS 1. (c). (b) 3. (b) 4. (a) 5. (d) 6. (c) 7. (c) 8. (c) 9. (a) 10. (c) 11. (b) 1. (a) 13. (b) 14. (c)

26 Subjective Problems m. acl.h O = g, H O = 4.96 g : % g ml 9. Pb(NO 3 ) = 3.3 g, NaNO 3 = 1.68 g 10. K CO 3 = 96%, Li CO 3 = 4% 11. SO 4 ion concentration = L 13. C H 6 = 0.66, C H 4 = g/mol mol L m gm