Syllabus : CONCEPTS. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18 CS 1

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1 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, CS S T I C H I E T RY Syllabus : C CNCEPTS In this chapter we will discuss the calculations based on chemical equations. It has been classified into two parts :. ole Concept. Equivalent Concept CA LE CNCEPT : In mole concept we deal with different types of relations like weight-weight, weight-volume, or volume-volume relationship between reactants or products of the reaction. Some basic formulas and definitions used in mole concept are as follows : ole is the unit representing numbers (atoms, molecules or ions). A mole also represents gram molecular mass or gram atomic mass. asses of atoms, molecules and nuclear particles are expressed in a specially defined unit, called atomic mass unit (u or amu). ne amu g / atom. olar mass of a substance is the mass of particles present in it. An empirical formula represents the simplest whole number ratio of various atoms present in a compound, whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. weight of substance Number of moles of a substance(n) atomic or molecular weight Also, The concept of atoms and molecules; Dalton s atomic theory; ole concept; Chemical formulae; Balanced chemical equations; Calculations (based on mole concept) involving common oxidationreduction, neutralization, and displacement reactions; Concentration in terms of mole fraction, molarity, molality and normality. Given number of molecules Number of moles of a substance(n) Avogadro number PV In gas phase reaction number of moles of a gas (n) RT

2 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, At STP/NTP one mole of any gas contains.4 L i.e. at 7 K and atm pressure. ole concept is based on balanced chemical reaction. If amounts of both reactants are given then find the limiting reagent. It is that reagent which is consumed completely in a irreversible chemical reaction. Class Discussion Problems :. Hydrogen reacts with nitrogen to produce ammonia according to this equation : H (g) + N (g) NH (g) Determine how much ammonia would be produced when 00.0 g of hydrogen react.. If 0.0 g of CaC is treated with 0.0 g of HCl, how many grams of C can be generated according to the following equation : CaC (s) + HCl(aq) CaCl (aq) + H (l) + C (g) Given : olar mass of (CaC 00 g mol, HCl 6.5 g mol, C 44.0 g mol ). Zinc and hydrochloric acid react according to the reaction : Zn(s) + HCl (aq) ZnCl (aq) + H (g) If 0.0 mol Zn are added to hydrochloric acid containing 0.5 mol HCl, how many moles of H are produced? 4. A compound contains 4.07% hydrogen,.47% carbon and 7.65% chlorine. Its molar mass is g. What are its empirical and molecular formulas? [At. wt. : H, C, Cl 5.5] 5. Find the number of atom in g of C. Home Work Problems : 6. In a reaction A + B AB. Identify the limiting reagent, if any, in the following reaction mixtures. (i) (ii) (iii) (iv) (v) 00 atoms of A + 00 molecules of B mol A + mol B 00 atoms of A + 00 molecules of B 5 mol A +.5 mol B.5 mol A + 5 mol B 7. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation : N (g) + H (g) NH (g). [At. wt. : N 4, H ] (i) Calculate the mass of ammonia produced if.00 0 g dinitrogen reacts with.00 0 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be the mass? 8. Which one of the following will have largest number of atoms? (i) g Au(s) (ii) g Na(s) (iii) g Li(s) (iv) g of Cl (g) [At. Wt. Au 97, Na, Li 7, Cl 5.5] [Answers : (). g () 8.80 g of C () 0.6 mol (4) CH Cl, C H 4 Cl (5) (6) (i) B is the limiting reagent (ii) A is the limiting reagent (iii) No limiting reagent (iv) B is the limiting reagent (v) A is limiting reagent (7) (i) g NH (ii) H will remain unreacted (iii) 57.4 g (8) g of Li has the largest number of atoms] C DIFFERENT WAYS F EXPRESSING THE CNCENTRATIN TERS : Important Definitions : mass of solute mass percent 00 mass of solution olarity() No. of moles of Vol. of solute, unit of molarity are mol/lit., or molar. solution in L CS

3 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, No. of gramequivalents of solute Normality (N), unit of normality are g-eq./lit., N or normal. Vol. of solution in L CS No. of gram equivalents of solute(neq ) Equivalent weight Weight of solute Equivalent weight olecular weight (or) Atomic weight (or) Ionic weight nfactor No.of molesof solute olality(m), unit of molality are mol/kg, m or molal. wt.of solvent olefraction( n A x A ). n A nb mass of solute ppm 0 mass of solution The relation between different concentration terms : 6. n eq n mol n-factor. n eq Normality Volume (L). Number of moles(n mol ) olarity Volume (L) 4. Normality olarity n-factor xd 6. xb 000 m ( x ) B A 000 m 000d (d density of solution in g/ml, molar mass of solute, x B and x A are mole fraction of solute and solvent respectively, A molar mass of solvent) Calculation of n Factor for Different Compounds :. Acids : n basicity H P 4 n H P n H P n H B n. Bases : n acidity of base e.g. Ammonia and all amines are monoacidic bases, NaH(n ), Na C (aq) n, NaHC (n ). Salt : (Which does not undergo redox reactions) n factor Total cationic or anionic charge, e.g. Na P 4 n, Ba (P 4 ) n 6 4. xidizing Agents or Reducing Agents : n factor change in oxidation number r number of electron lost or gained from one mole of the compound. Class Discussion Problems :. If the density of methanol is 0.79 kg L, what is its volume needed for making.5 L of its 0.5 solution?. A sample of drinking water was found to be severely contaminated with chloroform, CHCl supposed to be carcinogenic in nature. The level of contamination was 5 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.. Commercially available concentrated hydrochloric acid contains 8% HCl by mass. (a) What is the molarity of this solution? The density is.9 g ml. (b) What volume of concentrated HCl is required to make.00 L of 0.0 HCl?

4 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, ml of 0.50 Na solution is added to an aqueous solution of 5.00 g BaCl resulting in the formation of the white precipitate of insoluble Ba. How many moles and how many grams of Ba are formed? (At. wt. Ba 7, S, Na, 6, Cl 5.5) 5. Calculate the volume of.00 mol L aqueous sodium hydroxide that is neutralised by 00 ml of.00 mol L aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralisation reaction is : NaH(aq) + HCl (aq) NaCl (aq) + H (l) 6. What volume of 0.0 Na solution is required to prepare.0 L of a solution 0.40 in Na +? Home Work Problems : 7. What is the concentration of sugar (C H ) in mol L if its 0 g are dissolved in enough water to make a final volume up to L? 8. Calcium carbonate reacts with aqueous HCl to give CaCl and C according to the reaction CaC (s) + HCl(aq) CaCl (aq) + C (g) + H (l) What mass of CaC is required to react completely with 5 ml of 0.75 HCl? 9. The density of solution of NaCl is.5 g ml. Calculate the molality of the solution. C4 [Answers : () 5. ml () (i).5 0 % (ii) m () (a).8 (b) ml (4) 0.07 mol BaCl, 6.80 g Ba (5) 400 ml,.40 g (6). L (7) 0.09 mol L (8) 0.94 g CaC (9).79 m] EQUIVALENT CNCEPT It is based on law of equivalence which is explained as follows : Law of chemical equivalents : In a chemical reaction the equivalents of all the species (reactants or products) are equal to each other provided none of these compounds is in excess. N V N V (when normalities and volumes are given). If the number of equivalence of both the reactants are different then reactant with the lesser number of equivalence will be the limiting reagent. Application of equivalent concept : It is used in acid base titration, back titration and double titration, similarly in redox titration. Equivalent concepts can be used on all reaction whether they are balanced or not balanced but mole concept is used in solving the problems when the reactions are balanced. Basic principles of tirations : In volumetric analysis, a given amount (weight or volume) of an unknown substance is allowed to react with a known volume of a standard solution slowly. A chemical reaction takes place between the solute of an unknown substance and the solute of the standard solution. The completion of the reaction is indicated by the end point of the reaction, which is observed by the colour change either due to the indicator or due to the solute itself. Whether the reactions during the analysis are either between an acid and or base or between oxidising agent and reducing agent., the law of equivalence is used at end point. Following are the different important points regarding this process : (i) (ii) In case of acid base titration at the equivalence point (n eq )acid (n eq )base In case of redox titration (n eq )oxidant (n eq )reductant (iii) (iv) (v) (vi) If a given volume of solution is diluted then number of moles or number of equivalence of solute remains same but molarity or normality of the solution decreases. If a mixture contains more than one acids and is allowed to react completely with the base then at the equivalence point, (n eq ) acid + (neq) acid +... (n eq ) base Similarly if a mixture contains more than one oxidising agents then at equivalence point, (n eq ).A + (n eq ).A +... (n eq ) reducing agent. If it is a difficult to solve the problem through equivalence concept then use the mole concept. CS 4

5 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, Back titration : This is a method in which a substance is taken in excess and some part of it has to react with another substance and the remaining part has to be titrated against standard reagent. Double titration : This is a titration of specific compound using different indicators. Let us consider a solid mixture of NaH, Na C and inert impurities. When the solution containing NaH and Na C is titrated using phenolphthalein indicator following reaction takes place at the phenolphthalein end point NaH + HCl NaCl + H Na C + HCl NaHC + H Here, eq. of NaH eq.of Na C (n ) eq.of HCl When methyl orange is used, Na C is converted into NaCl + C + H Hence, eq. of NaH + eq. of Na C (n ) eq. of HCl TITRATIN F IXTURE F BASES WITH TW INDICATRS Every indicator has a working range Indicator ph range Behaving as Phenolphthalein 8 0 weak organic acid ethyl orange 4.4 weak organic base Thus methyl orange with lower ph range can indicate complete neutralisation of all types of bases. Extent of reaction of different bases with acid (HCl) using these two indicators summarised below Phenolphthalein ethyl range NaH 00% reaction is indicated 00 % reaction is indicated NaH + HCl NaCl + H NaH + HCl NaCl + H Na C 50% reaction upto NaHC 00% reaction is indicated stage is indicated Na C + HCl NaCl + H Na C + HCl NaHC + NaCl + C NaHC No reaction is indicated NaHC + HCl NaCl + H + C 00% reaction is indicated Redox titration : Class Discussion Problems :. What is the normality of 0.00 H P when it undergoes the following reaction? H P + H HP + H. What volumes of.0 N and.00 N HCl must be mixed to give.00 L of 6.00 N HCl?. Determine the normality of an H P 4 solution, 40.0 ml of which neutralized 0 ml of 0.5 N NaH. 4. A 48.4 ml sample of HCl solution requires.40 g of pure CaC for complete neutralization. Calculate the normality of the acid. (olecular wt. of CaC 00 g/mol) 5. In standardizing HCl,.5 ml was required to neutralize 5.0 ml of 0.00 N Na C solution. What is the normality of the HCl solution? How much water must be added to 00 ml of it to make it 0.00 N? Neglect volume changes. CS 5

6 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, 6. Given the unbalanced equation Kn 4 + KI + H K + n + I + H (a) How many g Kn 4 are needed to make 500 ml of 0.50 N solution? (b) How many g KI are needed to make 5.0 ml of 0.60 N solution? (At. wt. K 9, n 55, 6, I 7) 7. Determine the mass of Kn 4 required to make 80.0 ml of N/8 Kn 4 when the latter acts as an oxidizing agent in acid solution and n + is a product of the reaction. Home Work Problems : 8. Calculate the normality of each of the following solutions : (a) 7.88 g of HN per L solution (b) 6.5 g of Na C per L solution (if acidified to form C ). [At. wt. C, N 4, 6, Na ] 9. (a) What volume of 5.00 N H is required to neutralize a solution containing.50 g NaH? (b) How many g pure H are required? [olecular wt. of NaH 40, H 98] 0. Exactly 50.0 ml of a solution of Na C was titrated with 65.8 ml of.00 N HCl. If the density of the Na C solution is.5 g/ml, what percent Na C by weight does it contain? (At. wt. Na, C, 6). Calculate the mass of oxalic acid, H C 4, which can be oxidized to C by 00.0 ml of an n 4 solution 0.0 ml of which is capable of oxidizing 50.0 ml of.00 N I to I. (At. wt. H, C, 6). Find the equivalent weight of Kn 4 in the reaction n + + n 4 + H n + H + (unbalanced). How many g n is oxidized by.5 g Kn 4? (molar mass n 5 g) [Answers : () N () 0. L of conc. solution, L of dil. solution ().59 N (4) 0.5 N (5) 0. N, ml (6) (a).95 g (b).49 g KI (7) 0.6 g (8) 0.5 N, N (9) (a).5 ml (b).07 g (0) 6.8% Na C ().5 g ().79 g n ] C5 VLUE STRENGTH F H x volume of H means x litre of is liberated by volume of H on decomposition H H 68gm.4lit atstp Volume strength of H solution N (where N is the normality of the H solution Class Discussion Problems :. Calculate the strength of 0 V of H in terms of normality. In a 50 ml solution of H an excess of KI and dilute H were added. The I so liberated required 0 ml of 0. N Na S for complete reaction. Calculate the strength of H in grams per litre. [Answers : ().58 N () 0.68 g/litre] CS 6

7 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com,. Species Changed to Reactions n 4 (.A.) n + in acidic medium. n 4 (.A.) n in basic medium. n 4 (.A.) n 4 in neutral medium 4. Cr 7 (.A.) Cr + in acidic medium 5. n (.A.) n + in acidic medium n 4 + 8H + + 5e n + + 4H Estimation of Reaction Relation between.a. and R.A.. I I + Na S NaI + Na S 4 6 I + S I + S 4 6. Cu Cu + 4KI Cu I + K + I or Cu + + 4I Cu I + I white ppt.. CaCl CaCl + H Ca(H) + Cl Cl + KI KCl + I Cl + I Cl + I I I Na S Eq. wt. (Na S ) Cu I Na S Eq. wt. of Cu CaCl Cl I I Na S Eq. wt. of CaCl 4. n n + 4HCl (conc.) ncl + Cl + H Cl + KI KCl + I or n + 4H + + Cl n + + H + Cl Cl + I I + Cl Electron exchanged or change in.n. n 4 + e + H n + 4H n 4 + e + H n Cr 7 + 4H + + 6e Cr + + 7H n + 4H + + e n + + H 6. Cl (.A.) (in bleaching powder) Cl Cl + e Cl 7. Cu (.A.) Cu + Cu + + e Cu + (in iodometric titration) 8. S (R.A.) S 4 6 S S e (for two molecules) 9. H (.A.) H H + H + + e H 0. H (R.A.) H + H + + e (.N. of oxygen in H is () per atom). Fe + (R.A.) Fe + Fe + Fe + + e n Cl I I Na S Eq. wt. of n 5. I I + 5I + 6H + I + H I I 6I 6Na S Eq. wt. I Eq. wt. 5 6 CS 7

8 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, CS 8 6. H H + I + H + I + H H I I Na S Eq. wt. H 7. Cl Cl + I Cl + I Cl I I Na S Eq. wt. of Cl I + 6H + I + H + I Eq. wt. of 9. Cl Cl + I + H + H + Cl + I Cl I I Na S Eq. wt. of Cl 0. Cr 7 Cr 7 + 4H + + 6I I + Cr + + 7H. Fe + Fe + Fe + + e n 4 + 8H + + 5e n + + 4H. Fe + Fe + Fe + + e Cr 7 + 4H + + 6e Cr + + 7H. C 4 C 4 C + e n 4 + 8H + + 5e n + + 4H Cr 7 I 6I Eq. wt. of Cr 7 6 5Fe + n 4 Eq. Wt. Fe + Eq. Wt. n 4 5 6Fe + Cr 7 Eq. Wt. Cr 7 5C 4 n 4 6 Eq. wt. C 4 Eq. wt. n 4 /5 4. H H H e 5H n 4 5e + n 4 n + Eq. wt. H Eq. wt. n 4 5. As As + 5H As 4 + 0H + + 4e Eq. wt. As 5 4

9 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, CS 9 EXERCISE TYPE (SINGLE CRRECT CHICE TYPE). g of the carbonate of a metal was dissolved in 5 ml of N-HCl. The resulting liquid required 5 ml of N-NaH for neutralization. The eq. wt. of the metal carbonate is (a) 50 (b) 0 (c) 0 (d) None. If 0.5 mol of BaCl is mixed with 0.0 mol of Na P 4, the maximum amount of Ba (P 4 ) that can be formed is (a) 0.70 mol (b) 0.50 mol (c) 0.0 mol (d) 0.0 mol. The equivalent mass of n is half of its molar mass when it is converted to (a) n (b) n (c) n 4 (d) n 4 4. When one gram mol of Kn 4 reacts with HCl, the volume of chlorine liberated at NTP will be (a). litres (b).4 litres (c) 44.8 litres (d) 56.0 litres 5. 4 g of hydrogen peroxide is present in 0 ml of solution. This solution is called (a) 0 vol solution (b) 0 vol solution (c) 0 vol solution (d) vol solution 6. The number of moles of Kn 4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is (a) /5 (b) /5 (c) 4/5 (d) 7. [Na + ] in a solution prepared by mixing 0.00 ml of 0. NaCl with 70 ml of 0.5 Na is (a) 0.5 (b) 0.4 (c) 0.0 (d) ml of H solution (volume strength x) N required 0 ml of n4 solution in acidic 0.56 medium. Hence x is (a) 0.56 (b) 5.6 (c) 0. (d) olality of 8 H (d.8 gml ) is (a) 6 mol kg (b) 00 mol kg (c) 500 mol kg (d) 8 mol kg 0. The normality of an H P 4 solution 40.0 ml of which neutralized 0 ml of 0.5 N NaH is (a).00 N (b).59 N (c).00 N (d) 5.0 N. To neutralise completely 0 ml of 0. aqueous solution of phosphorus acid (H P ), the volume of 0. aqueous KH solution required is (a) 0 ml (b) 0 ml (c) 40 ml (d) 60 ml. 5. g of C is dissolved in 50 ml of N HCl. Unused acid required 00 ml of 0.5 N NaH. Hence equivalent weight of is (a) (b) (c) 4 (d). Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation H (aq.) H (l) + (g) under conditions where one mole of gas occupies 4 dm, 00 cm of X solution of H produces dm of X is thus (a).5 (b) (c) 0.5 (d) A solution contains Na C and NaHC. 0 ml of the solution required.5 ml of 0. H for neutralisation using phenolphthalein as indicator. ethyl orange in then added when a further.5 ml of 0. H was required. The amount of Na C and NaHC in litre of the solution is (a) 5. and 4. g (b). g and 6. g (c) 4. g and 5. g (d) 6. g and. g g of fuming H S 7 (leum) is diluted with water. This solution is completely neutralized by 6.7 ml of 0.4 N NaH. The percentage of free S in sample is (a) 0.6% (b) 40.6% (c) 0.6% (d) 50% g of a sample of Na C.xH were dissolved in water and the volume was made to 00 ml, 0 ml of this solution required 9.8 ml of N/0 HCl for complete neutralization. The value of x is (a) 7 (b) (c) (d) ml of Kn 4 oxidised 00 ml of H in acidic medium (when n 4 is reduced to n + ); volume of same Kn 4 required to oxidise 00 ml of H in basic medium (when n 4. is reduced to n ) will be

10 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, (a) (c) ml ml (b) (d) 500 ml 00 ml 8. For a given mixture of NaHC and Na C, volume of given HCl required is x ml with phenolphthalein indicator and further y ml required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHC is (a) x (b) x (c) y (d) (y x) 9. mol of ferric oxalate is oxidised by x mol of n 4 and also mol of ferrous oxalate is oxidised by y mol of n 4 in acidic medium. The ratio y x (a) : (b) : (c) : (d) : 0. mol of a mixture of Fe and Fe ( ) required 00 ml of Kn 4 solution is acidic medium. Hence mol fraction of Fe in the mixture is... (a) (c) 5 (b) (d) 5 TYPE- (CPREHENSIN TYPE) TYPE- (ATRIX-ATCH TYPE) Column-A Column-B (A) Total no. of electrons in (P) g methane are (B) No. of sulphate ions (Q) present in 50mL of 0. is CS 0 H solution are (C) 500cm of 0. molar NaCl (R) is added to 00cm of 0.5 molar AgN. Thus no. of ions of AgCl formed are (D) The no. of of Fe + ions (S).0 0 formed when excess of iron is treated with 5 ml of 0.04 N HCl TYPE-4 (T) (ULTIPLE CRRECT CHICE TYPE). Assuming complete dissociation of H as (H H + + ), the number of sulphate ions present in 50 ml of 0. N H solution are (a) (b) (c) 5 0 (d).0 0 mol. In the following redox reaction n 4 + 0Cl + 6H + n + + 5Cl + 8H. Pick up the correct statements. (a) (b) (c) (d) n 4 is reduced Cl is oxidising agent n 4 is an oxidising agent Cl is reduced. Which of the following is disproportionation? (a) Cu + Cu + Cu + (b) (c) (d) Cl + 6H Cl + 5Cl + H H S + 8 H + S Na + Cl NaCl 4. xidation number of C is zero in (a) CHCl (b) CH Cl (c) C 6 H 6 (d) C 5. Among the species given below which can act as oxidising as well as reducing agent? (a) S (b) S (c) H (d) H S TYPE-5 (Assertion-Reason Type) Each question contains STATEENT- (Assertion) and STATEENT- (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which NLY NE is correct. (A) (B) (C) (D) Statement- is True, Statement- is True; Statement- is a correct explanation for Statement- Statement- is True, Statement- is True; Statement- is NT a correct explanation for Statement- Statement- is True, Statement- is False Statement- is False, Statement- is True. STATEENT- :.4 L of ethane at N.T.P. contains one mole of hydrogen molecules. STATEENT- : ne mole of hydrogen

11 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com, molecules at N.T.P. occupies.4 L of volume.. STATEENT- : The masses of oxygen which combine with fixed mass of nitrogen in N, N, N bears a simple multiple ratio. STATEENT- : The combination according to law of multiple proportions.. STATEENT- : kg of Glauber s salt is dissolved in water to obtain dm of solution. The molarity of the resultant solution is STATEENT- : The volume in ml of 0.5 H needed to dissolve 0.5 g of copper (II) carbonate is 4. ml (At. Wt of Cu 6.5) 4. STATEENT- : Basicity of an acid can change in different reactions. STATEENT- : As the equivalent weight always remains same. 5. STATEENT- : If equal volume of C Kn 4 and C K Cr 7 solutions are allowed to oxidised Fe + to Fe + in acidic medium, then Fe + oxidised by Kn 4 is more then K Cr 7 of same concentrations. STATEENT- : Number of moles of electrons gained by mole of K Cr 7 is more than the mole of Kn 4 TYPE 6 (SUBJECTIVE TYPE PRBLES). If ml of.600 HCl and ml of.000 NaH are mixed, what are the molar concentrations of Na +, Cl, and H in the resulting solution? Assume a total volume of ml.. The acidic substance in vinegar is acetic acid CH CH. When 6.00 g of a certain vinegar was titrated with 0.00 NaH, 40. ml of base had to be added to reach the equivalence point. What percent by mass of this sample of vinegar is acetic acid?. ne gram of a mixture of CaC and gc gives 40 ml of C at N.T.P. Calculate the percentage composition of mixture (Ca 40, g 4, C, 6). 4. A solution contains 4 g of Na C and NaCl in 50 ml. 5 ml of this solution required 50 ml of N/0 HCl for complete neutralization. Calculate % composition of mixture. 5. In an Industrial process for producing acetic acid, oxygen gas is bubbled into acetaldehyde CH CH, containing manganese (II) acetate (catalyst) under pressure at 60 0 C. CH CH(l) + (g) CH CH(l) In a laboratory test of this reaction, 0.0 g CH CH CS and 0.0 g were put into a reaction vessel. (a) How many grams of acetic acid can be produced by this reaction from these amounts of reactants? (b) How many grams of the excess reactant remaining after the reaction is complete? 6. A 0.0 ml portion of (NH 4 ) solution was treated with excess NaH. The NH gas evolved was absorbed in ml of N HCl. To neutralize the remaining HCl,.50 ml of N NaH was required. What is the molar concentration of the (NH 4 )? How many g (NH 4 ) are in L solution? 7. A. g of a mixture containing H C 4.H and KHC 4.H and impurities of a neutral salt, consumed 8.9 ml of 0.5 N NaH for neutralization. n titrating with Kn 4 solution 0.4 g of the same substance needed.55 ml of 0.5 N Kn 4. Calculate the percentage composition of the substance. 8. A mixture of Fe and Fe is reacted with acidified Kn 4 solution having a concentration of 0.78, 00 ml of which was used. The solution then was treated with Zn dust which converted the Fe + of the solution to Fe +. The Fe + required 000 ml of 0. K Cr 7 solution. Find the mass % of Fe and Fe ml of 8 N nitric acid, 4.8 ml of 5 N HCl acid and a certain volume of 7 sulphuric acid are mixed together and made up to litre. Thirty ml of this acid mixture exactly neutralise 4.9 ml of sodium carbonate solution containing one gram of Na C. 0H in 00 ml of water. Calculate the mass in gram of the sulphate ions in solution. 0. A mixture of calcium carbonate and sodium chloride weighing.0 g was added to 00 ml of.0 N HCl. After the reaction had ceased the liquid was filtered, the residue washed and the filtrate was made up to 00 ml. 0 ml of this dilute solution required 5 ml of N/5 NaH for neutralization. Calculate the % of CaC in the mixture. (Ca 40).

12 Einstein Classes, Unit No. 0, 0, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -8 Ph. : 96905, 857, einsteinclasses00@gmail.com,.... a, b. a, c. a, b 4. b, c 5. a, c. D. A. B 4. C 5. D. a. d. b 4. d 5. a 6. a 7. d 8. d 9. c 0. b TYPE TYPE TYPE A-R ; B-P ; C-S ; D-T TYPE 4 TYPE 5 ANSWERS. c. a. a 4. a 5. c 6. c 7. b 8. d 9. a 0. a CS 5. (a) 7.4 gm (b).7 gm TYPE H, Cl,.Na %. CaC 6.5 %, gc 7.5% 4. Na C 66.5%, NaCl.75% , 9. g/l 7. KHC 4.H 80.9 %, H C 4.H 4.7 % 8. % Fe.4 %, % Fe % gm %