Q. No. 1 Which of the following contains equal to those in 1g Mg? (At. Wt. Mg = 4) 1 gm C 7 gm N 3 gm O None of these Correct Answer Explanation 1 1 Moles of Mg = = 4 Q. No. 1 Atoms = N A 7 1 Moles of N = = 8 4 1 NA No of atoms = NA = 4 1 If 1 moles of oxygen combine with Al to form Al O,the weight of Al used in the reaction is (Al = 7) 7 g 54 g 40.5 g 81 g Correct Answer Explanation 3 Al+ O AlO 3 3 mole mole 3 For mole of O, moles of Al required is mole at = 7 = 54gm Q. No. 3 Which has the highest mass? 50 g of iron 5 moles of N 0.1 mol atom of Ag 10 3 atoms of carbon Correct Answer Explanation 50 gm Fe 5 moles N = 5 8 = 140gm 1 mole Ag = 0.1 108 = 10.8 gm 3 3 10 1 10 atomof C = moles = moles 3 6.0 10 6.0 1 = 1= gm 6.0 3
Q. No. 4 The number of atoms present in 0.5 mole of nitrogen is same as the atoms in 1 g of C 64 g of S 8 g of O 48 g of Mg Explanation 0.5 mole N contains 0.5 N A molecules = 0.5 NA = NAatom This is same for 1 gm of C Q. No. 5 Which of the following weighs the least? g atom of N (at. wt. of N = 14) 3 3 10 atoms of C (at. wt of C = 1) 1 mole of S (at. wt. of S = 3) 7 g silver (at. wt. of Ag = 108) Correct Answer Explanation gm atom of N 8 gm 3 3 10 1 3 10 atomof C = = moles 3 6 10 1 = 1= 6gm 1 mole of S = 3 gm 7 gm Ag 3 Q. No. 6 If N A is Avogadro s number then number of valence electrons in 4. g of nitride ions (N 3 ) is.4 N A 4. N A 1.6 N A 3. N A Explanation 4. Moles = = 0.3 14 0.3 mole contains 0.3 N A ions One ion has 8 valence electron : total valence electron = 0.3N 8 =.4 N Q. No. 7 Hemoglobin contains 0.33 % of iron by weight. The molecular weight of hemoglobin is approximately 6700. The number of iron atoms (at. wt. of Fe = 56) present in one molecule of hemoglobin is 6 1 4 Explanation 100 gm of Haemoglobin contains 0.33 gm of iron. Let there are x atoms of iron present in one molecule. 100 1 Moles of Haemoglobin= = 6700 67 A A
x 0.33 Moles of Fe = = x 4 67 56 Q. No. 8 The number of molecules in 4.5 g of ammonia is about 3 1.0 10 3 1.5 10 3.0 10 3.5 10 Correct Answer Explanation 4.5 Moles = = 0.5 17 No. of molecules = 0.5 6.03 10 3 = 1.5 10 Q. No. 9 If 0% nitrogen is present in a compound, its minimum molecular weight can be 144 8 100 70 Explanation For minimum molecular weight 1 mole of compound must contain 1 mole of atom. Moles of compound = Moles of atom 100 0 = M= 70 M 14 Q. No. 10 The weight of molecule of the compound C 60 H 1 is -1 1.4 10 g -1 1.09 10 g 3 5.05 10 g 3 16.03 10 g Explanation (60 1+1 1) wt of one molecule = gm 6. 0 10 3 =1.4 10-1 gm Q. No. 11 Choose the wrong statement : 1 mole means 6.0 1 3 0 particles Molar mass is mass of one molecule Molar mass is mass of one mole of a substance Molar mass is molecular mass expressed in grams Correct Answer Explanation Molar mass is the mass of one mole of molecules. 3 Q. No. 1 Which among the following is the heaviest?
One mole of oxygen One molecule of sulphur trioxide 100 amu of uranium 44 g of carbon dioxide Explanation 1 mole O = 3 gm 1 molecule of SO = 80 1.66 10 100 amu of U= 100 1.66 10-7 kg 44 gm of CO 3 Q. No. 13 Rearrange the following I to IV in order of increasing masses and choose the correct answer [At. wt. of N = 14 u, O = 16 u, Cu = 63 u] I 1 molecule of oxygen II 1 atom of nitrogen III IV -10-7 1 10 mol molecule of oxygen -10 1 10 mol atom of copper II I III IV IV III II I II I II IV I II IV III Explanation I. 1 molecule of O = 3 a.m.u = 3 1.66 10-4 gm II. 1 atom of nitrogen = 14 a.m.u =14 1.66 10-4 gm -10 III. 10 mole o f O = 10-10 -10-10 3 gm IV. 10 mole of Cu = 10 63.5 gm II I III IV Q. No. 14 The number of moles of SO Cl in 13.5 g is 0.1 0. 0.3 0.4 Explanation 13.5 Moles = = 0.1 135 Q. No. 15 The largest number of molecules is in 36 g of water 8 g of carbon monoxide 46 g of ethyl alcohol 54 g of nitrogen pentoxide Explanation 36 36 gm water = moles = moles 18 No. of molecules = N A
8 8 gmco= =1 mole 8 No. of molecules = 1 N A 46 46 gm of CH3CH OH= =1mole 46 no of molecules = 1 N A 54 54 gmof NO 5 = moles 108 1 1 = mole = N molecules A Q. No. 16 Which of the following contains maximum number of atoms? 1 6.03 10 molecules of CO.4 L of CO at STP 0.44 g of CO None of these Correct Answer Explanation 1 6.03 10 molecules of O 1 6.03 10 - = moles =10 moles 3 6.0 10 - No of atom=10 N 3 atom.4.4 lit of CO = =1 mole.4 Atom = 1 NA 3 = 3N Aatom 0.44 0.44 gmco = moles = 0.01 moles 44 atom = 0.01 N 3 atom A A Q. No. 17 If 0.5 mol of BaCl is mixed with 0. mol of Na 3 PO 4, the maximum number of mole of Ba 3 (PO 4 ) that can be formed is 0.7 0.5 0.30 0.10 Explanation 3 BaCl +Na PO Ba(PO) + 6NaCl 3 4 3 4 0.5 mole 0. mole 0.5 mole BaCl requir es = 0.5 mole Na PO 3 3 = 0.33 mole Na 3 PO 4 Na PO is the limiting reactant 3 4 3 4 4 moles of Ba ( PO ) 0. 1 mole Q. No. 18 One mole of a mixture of CO and CO requires exactly 0 gram of NaOH in solution for
complete conversion of all the CO into Na CO 3. How many moles more of NaOH would it require for conversion into Na CO 3 if the mixture (one mole) is completely oxidised to CO 0. 0.5 0.4 1.5 Explanation Total moles of mixture = 1 mole Let moles of CO = x moles of CO =(1 -) x CO + NaOH NaCO 3 +HO 0 1 Moles of NaOH= = mole 40 1 mole of CO requires mole of NaOH ()( 1 - x mole of COrequires 1 - x) mole of NaOH 1 (1-) x= 3 x = 4 3 3 1 Moles of CO=, moles of CO =1- x =1- = 4 4 4 On oxidation only CO converts its CO 3 Moles of COformed= 4 Total moles of CO = 1 mole Moles of NaOH required = mole Extra moles of NaOH required = - 0.5 = 1.5 Q. No. 19 The number of water molecules present in a drop of water (volume = 0.0018 ml) at room temperature is (density of H O = 1 g/ml) 19 6.03 10 18 1.084 10 17 4.84 10 3 6.03 10 Explanation Mass of water = v = 1 0.0018 = 0.0018 gm 0.0018 Moles = = 0.0001 18 No of molecules = 0.0001 6.0 10 3 19 = 6.0 10 Q. No. 0 What is the weight of oxygen required for the complete combustion of.8 kg of
ethylene?.8 kg 6.4 kg 9.6 kg 96 kg Explanation C H + 3O CO + H O 4 3.8 10 Moles of CH 4 = = 100 moles 8 Moles of O required = 300 moles wt of oxygen = 300 3 gm = 9.6 kg Q. No. 1 A sample of pure calcium weighing 1.35 g was quantitatively converted to 1.88 g of pure calcium oxide. Atomic mass of calcium would be: 0 40 16 35.5 Correct Answer Explanation 1 Ca+ O CaO 1.35 Moles of Ca= M 1.35 Moles of CaO = M 1.88 Moles of CaO = 56 1.35 1.88 = M= 40 M 56 Q. No. 30 g of magnesium and 30 g of oxygen are reacted, then the residual mixture contains 60 g of Magnesium oxide only 40 g of Magnesium oxide and 0 g of oxygen 45 g of Magnesium oxide and 15 g of oxygen 50 g of Magnesium oxide and 10 g of oxygen Explanation Mg + O MgO 30 gm 30 gm 30 Moles of Mg = =1.5 4 30 Moles of O = = 0.9375 3 1.5 1.5 moles of Mg requires moles of O = 0.65 moles Mg is the limiting reactant Moles of MgO formed = 1.5 Mass of MgO = 1.5 40 gm = 50 gm
Moles of O left = 0.9375-0.65 = 0.315 wt of O = 0.315 3gm = 10 gm Q. No. 3 Silicon carbide, is produced by heating SiO and C to high temperatures according to the equation : SiO(s) + 3C(s) SiC(s) + CO(g) How many grams of SiC could be formed by reacting.00g of SiO and.0 g of C? 1.33.56 3.59 4.0 Explanation SiO + 3C (s) SiC + CO gm gm Q. No. 4 Given the reaction Moles of SiO = = 0.033 moles 60 Moles of C = = 0.166 moles 1 C is in excess and the limiting reactant is SiO Moles of SiC formed = 0.033 wt of SiC formed = 0.033 40 gm = 1.3 gm Pb(NO)(aq) 3 + KI PbI(s) + KNO(aq) 3 What is the mass of PbI that will precipitate if 10. g of Pb(NO 3 ) is mixed with 5.73 g of KI in a sufficient quantity of H O?.06 g 4.13 g 7.96 g 15.9 g Explanation Pb(NO) + KI PbI +KNO 3 3 10. gm 5.73 gm 10. Moles of Pb(NO) 3 = = 0.0307 33 5.73 Moles of KI = = 0.0345 166 L.R is K.I Moles of PbI formed = 0.017 wt of PbI = 0.017 46 = 7.946 gm Q. No. 5 If 9 moles of O and 14 moles of N are placed in a container and allowed to react
according to the equation : 3O + N NO 3 The reaction proceeds until 3 moles of O remain, how many moles of N O 3 are present at that instant? 6 3 4 1 Explanation 3O + N N O3 9 moles 14 moles O is the limiting reactant ideally. When 3 moles of O remains at that instant moles of O reacted = 6 moles moles of NO 3 formed= 6 = 4mole 3 Q. No. 6 Iron (III) oxide can be reduced with CO to fo rm metallic iron as described by unbalanced chemical reaction FeO 3 + CO Fe + CO The number of moles of CO required to form one mole of Fe from its oxide is 1 1.5 3 Correct Answer Explanation FeO 3 + 3CO Fe + 3CO mole of Fe is formed from 3 moles of CO 3 1 mole of Fe is formed from moles of CO Q. No. 7 The mass of CaO that shall be obtained by heating 0 kg of 90 % pure lime-stone (CaCO 3 ) is 11. kg 8.4 kg 10.08 kg 16.8 kg Explanation CaCO CaO + CO(g) 3 0,000 90 wt of pure CaCO 3 = 100 = 18000 gm 18000 Moles of CaCO 3 = =180 moles 100 moles of CaO = 180 wt of CaO = 180 56gm = 10.08 kg Q. No. 8 If potassium chlorate is 80 % pure, then 48 g of oxygen would be produced from
(atomic mass of K = 39) 153.1 g of KClO 3 1.5 g of KClO 3 45 g of KClO 3 98.0 g of KClO 3 Correct Answer Explanation KClO 3(s) KCl +3O(g) 48 Moles of O = =1.5 3 3 moles of O is formed from mole KClO 3 1.5 moles of O is form ed fro m 1.5 moles KClO 3 = 1 mole KClO 3 3 Q. No. 9 Antimony reacts with sulphur according to the equation Sb(s) + 3S(s) Sb S(s) 3 The molar mass of Sb S 3 is 340 g mol 1 What is the percentage yield for a reaction in which 1.40 g of Sb S 3 is obtained from 1.73 g of antimony and a slight excess of sulphur? 80.9% 58.0% 40.5% 9.0% Correct Answer Explanation Sb(s) + 3S Sb S 3 1.73 gm 1.73 Moles of Sb = = 0.0141 1 Moles of Sb S 3 ideally formed 0.0141 = = 0.00705 Ideal wt of SbS 3 = 0.00705 [340] gm =.397 But actual wt of Sb S 3 = 1.40 gm 1.40 % yield= 100 = 58%.397 Q. No. 30 NH 3 is produced according to the following reaction : N(g)+3H(g) NH(g) 3 In an experiment 0.5 mol of NH 3 is formed when 0.5 mol of N is reacted with 0.5 mol of H. What is % yield? 75% 50% 33% 5% Explanation N(g) +3H(g) NH(g) 3 0.5 mole 0.5 mole
H is the limiting reactant ideally the moles of NH 3 formed = 0.5= 0.34 3 Actual moles of NH 3 formed = 0.5 0.5 yield= 100 = 75 0.34 Q. No. 31 What is the weight % sulphuric acid in an aqueous solution which is 0.50 M in sulphuric acid? The specify gravity of the solution is 1.07. 4.77% 5.67% 9.53%.0% Explanation Molarity = 0.50 M 50 moles of HSO4 is present in 1 lit of solution i.e. 100 ml of solution Mass of solution = V = 1.07 1000 = 1070 gm Mass of solute = 0.50 98 = 49. 196 gm 49.096 wt % = 100 = 4.77% 1070 Q. No. 3 Mole fraction of ethanol in ethanol-water mixture is 0.5. Hence, percentage concentration of ethanol (C H 6 O) by weight of mixture is 5 75 46 54 Explanation X = 0.5= 1 CH5 OH 4 1 mole of CH5OH is present in 4 mole of solution Moles of solvent = 4-1 = 3 mole wt of C H 5 OH = 46 gm wt of solven t = 3 18 = 54 gm Total wt of solution = 46 + 54 = 100 gm % 46 by wt = 100 = 46 100 % Q. No. 33 A molal solution is one that contains one mole of a solute in 1000 g of the solvent One litre of the solvent One litre of the solution.4 litres of the solution Explanation In a molal solution one mole of solute is present in 1000 gm of solvent.
Q. No. 34 An aqueous solution of ethanol has density 1.05 g/ml and it is M. What is the molality of this solution? 1.79.143 1.951 None of these Correct Answer Explanation M C H 5 OH moles of C H 5 OH is present in 1 lit or 1000 ml of solution. Mass of solution = V = 1.05 1000 = 105 gm Mass of solute = 46 gm = 9 gm Mass of solvent = 105-9 = 933 gm 1000 Molality = =.14 933 Q. No. 35 What volume of 0.4 M FeCl 3.6H O will contain 600 mg of Fe 3+? 49.85 ml 6.78 ml 147.55 ml 87.65 ml Correct Answer Explanation Let volume V lit moles of solute = 0.4 V 3+ moles of Fe = 0.4V 3+ 600 moles of Fe = 0.4V 56 = 1000 V = 6.78 ml Q. No. 36 A sample of H SO 4 (density 1.8 g/ml) is 90% by weight. What is the volume of the acid that has to be used to make 1 litre of 0. M H SO 4? 16 ml 10 ml 1 ml 18 ml Explanation 100 gm of solution contain 90 gm of H SO 4 100 1 Vol. of solution= lit 1.8 1000 1 = lit 18 90 Moles of HSO 4 = 98 90 18 Molarity = =16.53 98 1 Let volume of solution = V lit 16.53 V = 1 0. [Moles of solute will be same] V 1 ml
Q. No. 37 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 1.45 1.64 1.88 1. Explanation 3.60 moles of H SO 4 is present in 100 ml of solution. Let density = d gm/ml M ass of solu tion = d 1000 Mass of solut e = 3.60 98 gm = 35.8 gm 35.8 by wt = 100 =9 1000d d = 1. gm/ml Q. No. 38 An antifreeze mixture contains 40% ethylene glycol (C H 6 O ) by weight in the aqueous solution. If the density of this solution is 1.05 g ml, what is the molar concentration? 6.77 M 6.45 M 0.0017 M 16.9 M Explanation 100 gm of solution contain 40 gm ethylene glycol (C H 6 O ). 100 Vol. of solution= ml 1.05 1 = lit 10.5 40 Moles of solute = 6 40 10.5 Molarity = = 6.77M 6 1 Q. No. 39 - What is the molarity of SO4 ion in aqueous solution that contain 34. ppm of Al (SO 4 ) 3? (Assume complete dissociation and density of solution 1 g/ml) 3 10-4 M 10-4 M 10-4 M None of these Explanation 34. gm of Al (SO 4 ) 3 is present in 10 6 gm of solution 10 Vol. of solution = ml 1 = 10 3 lit 34. moles of solute = = 0.1 mole 34 3+ - 4 3 4 0.1 mole 0.3 mole Al(SO) Al + 3SO 6
- 0.3 Molarity of SO 4 = 10 3 = 3 10-4 M Q. No. 40 The mole fraction of a given sample of I in C 6 H 6 is 0.. The molality of I in C 6 H 6 is 0.3 3. 0.03 0.48 Correct Answer Explanation 1 X I = 0. = = 10 5 i.e. 1 mole of I is present in 5 mole of solution. mole of C6H 6 = 5-1 = 4 wt of C6H 6 = 4 78 gm = 31 gm 1 1000 molality = = 3.0 31 Q. No. 41 In which mode of expression, the concentration of a solution remains independent of temperature? Molarity Normality Formality Molality Explanation Molality in values the wt. of solvent which does not changes with temperature. Q. No. 4 With increase of temperature, which of these changes? Molality Weight fraction of solute Fraction of solute present in unit volume of water Mole fraction Explanation With increase of temperature volume changes. Q. No. 43 Molarity and Normality changes with temperature because they involve: Moles Equivalents Weights Volumes Explanation But molarity and normality changes with temperature because they involve volume and volume changes with temperature. Q. No. 44 When 500.0 ml of 1.0 M LaCl 3 and 3.0 M NaCl are mixed. What is molarity of Cl ion? 4.0 M
3.0 M.0 M 1.5 M Correct Answer Explanation M illim oles of LaCl3 = 500 1 = 500 - m. moles of Cl = 1500 M illim oles of N acl = 500 3 = 1500 - m. moles of Cl = 1500 Total m. moles of Cl - = 3000 Total volume = 100 ml - 3000 Molarity of Cl = = 3 1000 Q. No. 45 When 50 ml of.00 M HCl, 100 ml of 1.00 M HCl and 100 ml of 0.500 M HCl are mixed together, the resulting HCl concentration of the solution is 0.5 M 1.00 M 3.50 M 6.5 M Correct Answer Explanation 50 +100 1+100 0.5 M net = 50+100+100 = 1 M Q. No. 46 A sample of H SO 4 (density 1.8 g ml 1 ) is 90% by weight. What is the volume of the acid that has to be used to make 1 L of 0. M H SO 4? 16 ml 18 ml 1 ml 10 ml Explanation 100 gm sample contain 90 gm of H SO 4 100 1 Vol. of sampl e= ml= lit 1.8 18 90 18 Molarity of solution= =16.53 98 1 Let the vol. of solution taken = V lit Moles of acid = V 16.53 = 1 0. or 0.1 ml V=0.01 Q. No. 47 What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO 3 and 0.1 M NaCl are mixed together?
0.1 M 0. M 0.05 M 0.5 M Explanation AgNO +NaCl AgCl+NaNO 3 3 0.1 V 0.1 V 0 0 0.1V 0.1V 0.1V [NO ]= = 0.05M V - 3 Q. No. 48 How many grams of NaBr could be formed if 14. g of NaI are reacted with 40.0 ml of a 0.800 M Br? NaI+Br NaBr+ I 3.30 4.80 6.59 9.75 Explanation NaI + Br NaBr+ I 14. gm 40 ml of 0.8M 14. Moles of NaI= = 0.0946 150 0.8 40 Moles of Br = = 0.03 1000 0.0946 0.0946 moles of NaI will be reacting with mole of Br i.e. 0.0473 moles Br is limiting reactant Moles of NaBr formed = 0.3 = 0.064 w t of NaBr = 0.064 103 = 6.59 Q. No. 49 If AgBr is assumed to be completely insoluble, What mass of AgBr precipitates when 30.0 ml of a 0.500 mol/l solution of AgNO 3 is added to 50.0 ml of an 0.400 mol/l solution of NaBr? 3.76 g 1.8 g.8 g 3.76 kg Explanation AgNO +NaBr Ag Br+NaNO 3 3 30ml, 50ml, 0.500M 0.4M M illimoles of AgNO3 = 30 0. 5 = 15 Millimoles of NaBr =50 0.4 = 0 Limiting reactant is AgNO 3 Millimoles of AgBr = 15
15 M ass of AgBr = [188] gm 1000 =.8 Q. No. 50 In a titration, 15.0 cm 3 of 0.100 M HCl neutralizes 30.0 cm 3 of Ca(OH). What is the molarity of Ca(OH) solution? 0. 015 0.050 0.0500 0.00 Correct Answer Explanation Milli eq. of HCl = mlli. Eq of Co(OH) (M V 'n'factor) =(M V 'r'factor) 0.1 15 1=30 M M = 0.05 HCl Ca(OH) Q. No. 51 10 ml of 1 M BaCl solution and 5 ml 0.5 M K SO 4 are mixed together to precipitate out BaSO 4. The amount of BaSO 4 precipitate will be 0.005 mol 0.0005 mol 0.05 mol 0.005 mol Explanation BaCl +KSO4 BaSO 4 + KCl m. moles of BaCl = 10 m. moles of K SO 4 =.5 K SO is the limiting reactant 4 m. moles of BaSO 4 =.5 = 0.005 mole Q. No. 5 Mg of a substance when vaporised occupy a volume of 5.6 litre at NTP. The molecular mass of the substance will be: M M 3 M 4 M Explanation Mass of 5.6 lit at N.T.P is M gm Mass of.4 lit at N.T.P is M.4 5.6 = 4 M Q. No. 53 Number of molecules in 1 litre of oxygen at NTP is : 3 6.0 10 3 3 6.0 10.4 3.4
3.4 Correct Answer Explanation 1 Moles of O =.4 1 No. of molecular = 6.0 10 3.4 Q. No. 54 The number of molecules in 89.6 litre of a gas at NTP are : 3 6.0 10 6.0 10 3 6.0 10 4 6.0 10 Explanation 89.6 Moles = = 4.4 No. of molecular = 4 6.0 10 3 Q. No. 55 The mass of 11 cm 3 of CH 4 gas at STP is 0.16 g 0.8 g 0.08 g 1.6 g Explanation 11 1 Moles = = 400 00 1 Mass = 16 = 0.08 gm 00 3 3 3 Q. No. 56 An oxide of metal (M) has 40% by mass of oxygen. Metal M has atomic mass of 4. The empirical formula of the oxide is M O M O 3 MO M 3 O 4 Explanation Let the valency of metal = x 100 gm of oxide contain 40 gm of oxygen wt of metal = 100-40 = 60 gm 40 gm of oxygen combines with 60 gm of metal 60 8 gm of oxygen combines with 8 gm of metal 40 Equivalent wt of metal = 1 gm 4 Eq wt = 1 = Valency = Valency Formula = MO
Q. No. 57 What is the empirical formula of a compound composed of O and Mn in equal weight ratio? MnO MnO Mn O 3 Mn O 7 Explanation The wt. of Mn and O are equal x Moles of Mn= 55 Moles of O= 16 x x x Mole ratio of Mn : O= ; 55 16 It empirical formula = Mn O 7 Q. No. 58 Determine the empirical formula of Kelvar, used in making bullet proof vests, is 70.6% C, 4.% H, 11.8% N and 13.4% O : C 7 H 5 NO C 7 H 5 N O C 7 H 9 NO C 7 H 5 NO Explanation Given : C = 70.6%, H = 4.%, N = 11.8%, O =13.4% Moles C 70.6 = 5.88 1 H 4. = 4. 1 N 11.8 = 0.84 14 Simplest of ratio 5.88 = 7.0 0.8375 4. =5.01 0.8375 0.84 =1.005 0.8375 O 13.4 = 0.8375 16 Empirical formula C H NO 7 5 0.8375 =1 0.8375 Q. No. 59 A compound contains atoms of three elements A, B and C. If the oxidation number of A is +, B is +5 and C is -, the possible formula of the compound is: A(BC 3 ) A 3 (BC 4 ) A 3 (B 4 C) ABC Explanation The sum of oxidation number of all the element, in a compound is equal to zero. In (a) i.e. A(BC) 3,sum of oxidation numbe r is + + [5-6] = 0 In (b) i.e. A 3 (BC 4 ) sum of oxidation = 0 Q. No. 60 The carbonate of a metal is isomorphous (similar formula) with magnesium carbonate
and contains 6.091% of carbon. The atomic weight of metal is 4 56 137 60 Explanation The formula of carbonate of metal = MCO 3 % 1 C = 100 =6.091 M+60 M = 137 Q. No. 61 The Ew of an element is 13. It forms an acidic oxide which with KOH forms a salt isomorphous with K SO 4. The atomic weight of element is 13 6 5 78 Explanation Formula of salt will be M SO 4 Valency of element is 1 M Equivalent wt = 1 M 13 = 1 M = 13 Q. No. 6 A hydrate of Na SO 3 losses.% of H O by mass on strong heating. The hydrate is Na SO 3.4H O Na SO 3.6H O Na SO 3.H O Na SO 3.H O Explanation Let the hydrate is Na SO 3. xh O 100 gm of hydrate looses. gm of H O on heating 3 3 100 gm. Na SO.xH O Na SO + xh O 100 Moles of hydrate = 16+18x 100x It will gives moles of HO 16+18x. mole of HO= 18 100x. = x = 16+18x 18 Na SO.H O 3 Q. No. 63 One of the following combinations illustrate law of reciprocal proportions N O 3, N O 4, N O 5 NaCl, NaBr, Nal CS, CO, SO
PH 3, P O 3, P O 5 Explanation C S(in CS) 1 gm 64 gm C O(in CO) 1 gm 3 gm wt ratio of S : O = 64 : 3 = : 1 when S and O combines directly then they form SO in which weight ratio of S : O is 3 : 3 = 1 : 1. This illustrate the law of reciprocal proportion. Q. No. 64 If water samples are taken from sea, river, clouds, lake or snow, they will be found to contain H and O in the approximate ratio 1 : 8. This indicates the law of Multiple proportion Definite proportion Reciprocal proportions None of these Correct Answer Explanation Any chemical compound always contain fixed ratio by wt of element in it no matter they are prepared from any source or by any chemical method. This is law of definite proportion. Q. No. 65 The law of multiple proportion is illustrated by Carbon monoxide and carbon dioxide Potassium bromide and potassium chloride Water and heavy water Calcium hydroxide and barium hydroxide Explanation When two elements combine to for more than one compound then with a fixed wt. of one element the wt. ratio of other elements combining bears a simple whole no multiple. This is the law of multiple proportion. Q. No. 66 The percentage of copper and oxygen in samples of CuO obtained by different methods were found to be the same. This illustrates the law of Constant proportions Conservation of mass Multiple proportions Reciprocal proportions Explanation Q. No. 67 Two samples of lead oxide were separately reduced to metallic lead by heating in a current of hydrogen. The weight of lead from one oxide was half the weight of lead obtained from the other oxide. The data illustrates. Law of reciprocal proportions Law of constant proportions Law of multiple proportions Law of equivalent proportions Explanation With a fixed mass of oxygen the wt of lead combining bears a simple whole no multiple. This is the law of multiple proportion.
Q. No. 68 One part of an element A combines with two parts of another element B. Six part of the element C combine with four parts of the elements B. If A and C combine together the ratio of their weights will be governed by Law of definite proportions Law of multiple proportions Law of reciprocal proportions Law of conservation of mass Explanation If A and B combine directly to form a compound and if C and B combine to form and compound if A and C combine directly to form a 3 rd compound then wt ratio of A and C will be governed by law of reciprocal proportion. Q. No. 69 n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as follows. X + Y = R + s The relation which can be established in the amounts of the reactants and the products will be n - m = p - q n + m = p + q n = m p = q Correct Answer Explanation X + Y = R + S n gm m gm p gm g gm Total wt of reactants = Total wt of products n + m = p + q Q. No. 70 Which one is the best example of law of conservation of mass? 6 g of carbon is heated in vacuum, there is no change in mass 6 g of carbon combines with 16 g of oxygen to form g of CO 6 g water is completely converted into steam A sample of air is heated at constant pressure when its volume increases but there is no change in mass Correct Answer Explanation During any chemical reaction total mass is conserved. Q. No. 71 SO gas was prepared by (i) burning sulphur in oxygen, (ii) reacting sodium sulphite with dilute H SO 4 and (iii) heating copper with conc. H SO 4. It was found that in each case sulphur and oxygen combined in the ratio of 1 : 1. The data illustrates the law of : Conservation of mass Multiple proportions Constant proportions Reciprocal proportions Explanation Q. No. 7 A sample of CaCO 3 has Ca = 40% C = 1% and O = 48% If the law of constant proportions is true, then the mass of Ca in 5 g of CaCO 3 from another source will be:.0 g 0. g
0.0 g 0.0 g Correct Answer Explanation Any chemical compound always has same % of elements no matter it is obtained from any source. 40 Mass of Ca = 5 100 = gm Q. No. 73 H S contains 5.88% hydrogen, H O contains 11.11% hydrogen while SO contains 50% sulphur. These figures illustrate the law of: Conservation of mass Constant proportions Multiple proportions Reciprocal proportions Explanation In H S, H = 5.88 % S = 94.1 In H O, H = 11.11 %,O = 88.89 wt ration S : O = 1 : 1 In SO, S = 50,O = 50 wt ratio of S : O = 1 : 1 This illustrate the law of reciprocal proportion. Q. No. 74 Hydrogen combines with chlorine to form HCl. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as: 3 : 35.5 35.5 : 3 1 : 1 3 : 1 Explanation In HCl, wt ratio of H : Cl = 1 : 35.5 In NaOH, wt ratio of H : Na = 1 : 3 If Na and Cl combine directly then wt ratio = 3 : 35.5 Q. No. 75 x g of Ag was dissolved in HNO 3 and the solution was treated with excess of NaCl when.87 g of AgCl was precipitated. The value of x is 1.08 g.16 g.70 g 1.6 g Correct Answer Explanation Ag + HNO AgNO NaCl 3 3 AgCl Mole ratio is 1 : 1 Moles of Ag = Moles of AgCl x.87 = 108 143.5 x =.16 gm
Q. No. 76 A 1.50 g sample of an ore containing silver was dissolved, and all the Ag + was converted to 0.15 g Ag S. What was the percentage of silver in the ore? 14.3% 10.8% 8.7% 70% Explanation 0.15 Moles of AgS = 48 0.15 moles of Ag in AgS = 48 = 0.001 wt of Ag = 0.001 108 = 0.1088 0.1088 % of Ag = 100 = 7.% 1.50 Q. No. 77 NaOH is formed according to the reaction 1 Na + O NaO NaO + HO NaOH To make 4 g of NaOH, Na required is 4.6 g 4.0 g.3 g 0.3 g Explanation 4 Moles of NaOH= = 0.1 mole 40 Mole ratio of Na and NaOH is 1 : 1 Moles of Na = 0.1 mole wt of Na =.3 gm Q. No. 78 H 3PO 4 + 3Ca(OH) Ca3 ( PO 4) + 6 H O Equivalent weight of H3PO 4in this reaction is 98 49 3.66 4.5 Explanation 98 Equal= = 3.66 or H3PO 4 has lost all root it 3 Q. No. 79 The Ew of H 3 PO 4 in the reaction is Ca(OH) + H3 PO4 CaHPO 4 + H O (Ca = 40, P = 31, O = 16) 49
98 3.66 147 Explanation Ca(OH) +H PO CaHPO +H O 3 4 4 In the reaction H 3 PO 4 has lost H its 'n' factor = M 98 Eq.wt = = = 49 Q. No. 80 What weight of a metal of equivalent weight 1 will give 0.475 g of its chloride? 0.1 g 0.4 g 0.36 g 0.48 g Explanation x M + Cl MClX n factor of metal = valency of metal Let the wt of metal = y gm Gm - eq of metal = gm - eq of MCl X y 0.475 = M M + x(35.5) x x M =1 x On solving y = 0.1 gm Q. No. 81 How many grams of phosphoric acid would be needed to neutralise 100 g of magnesium hydroxide? (The molecular weights are : H 3 PO 4 = 98 and Mg (OH) = 58.3) 66.7 g 5 g 11 g 168 g Explanation Gm - eq of H 3 PO 4 = gm - eq of Mg(OH) x 100 3 = 98 58 x = 11 gm Q. No. 8 0.116 g of C 4 H 4 O 4 (A) is neutralised by 0.074 g of Ca(OH). Hence, protonic hydrogen 1 () Hin (A) wil l be
3 4 Correct Answer Explanation Gm - equivalents of C 4 H 4 O 4 = gm - eq of Ca(OH) G m - eq of C 4H4O 4 = moles 'n'factor 0.116 =() x 116 [where x is the no of H + released] 0.074 Gm - eq of Ca(OH ) = 74 0.116 0.074 x = x = 116 74 Q. No. 83 4. g of metallic carbonate MCO 3 was heated in a hard glass tube and CO evolved was found to have 110 ml of volume at STP. The Ew of the metal is 1 4 18 15 Explanation MCO MO + CO 3 4. gm 110 ml of S.T.P. 110 Moles of CO = = 0.05 400 4. Moles of MCO 3 = M + 60 4. = 0.05 M=4 M + 60 4 Eq. wt of metal= =1 Q. No. 84 1.0 g of a monobasic acid when completely aceted upon Mg gave 1.301 g of anhydrous Mg salt. Equivalent weight of acid is 35.54 36.54 17.77 18.7 Correct Answer Explanation Let the monobasic acid = HA Mg + HA MgA + H 1 gm 1.301 gm 1 1.301 Moles of HA =,Moles of MgA = A+1 4+A As moles of HA gives 1 mole of MgA 1 1 1.301 moles of HA gives = A + 1 (A + 1) 4 + A
A = 35.54 Molecular wt = 35.54 + 1 = 36.54 Eq. wt = 36.54 /1 Q. No. 85 0.1 g of metal combines with 46.6 ml of oxygen at STP. The equivalent weight of metal is 1 4 6 36 Explanation 46.6 Moles ofoxygen= = 0.000 400 wt of oxygen = 0.000 3 = 0.064 gm 0.064 gm of O combines with 0.1 gm of metal 0.1 8 gm of O combines with= 8 =1 gm 0.064 Q. No. 86 When 100 ml of 1 M NaOH solution and 10 ml of 10 N H SO 4 solution are mixed together, the resulting solution will be : Alkaline Acidic Strongly acidic Neutral Explanation milli - eq of NaOH = 1 1 100 = 100 m. eq of HS O 4 = 10 10 = 100 Solution is neutral Q. No. 87 Normality of 0.74 g Ca(OH) in 5 ml solution is 8 N 4 N 0.4 N N Correct Answer Explanation Gm - equivalents of Ca(OH) 0.74 1 = = 74 50 1 1000 N= = 4N 50 5 Q. No. 88 Normality of a M sulphuric acid is 4 N N/ N/4 Correct Answer
Explanation N = M 'n' factor = = 4N Q. No. 89 1 L of a normal solution is diluted to 000 ml. The resulting normality is : N/ N/4 N N Explanation Gm - eq. of so lution = 1 1 = 1 1 1000 New normality = 000 1 N = i.e Q. No. 90 What volume of 0.3 N solution contains 3.17 milliequivalent of solute? 137 ml 13.7 ml 7.3 ml 1.7 ml Correct Answer Explanation Milliequivalents of solute = N V = 0.3 V (ml) = 3.17 V = 13.7 ml Q. No. 91 1 L solution of NaOH contains 4.0 g of it. What shall be the difference between molarity and the normality? 0.10 Zero 0.05 0.0 Correct Answer Explanation For NaOH n factor = 1 N and M are same Difference is zero Q. No. 9 100 ml of 0.3 N HCl is mixed with 00 ml of 0.6 N H SO 4. The final normality of the resulting solution will be 0.1 N 0. N 0.3 N 0.5 N
Explanation m. eq of HCl = 30 m. eq of HSO 4 = 00 0.6 = 10 Total gm - eq = 150 Total volume = 300 ml 150 1 N Final = = = 0.5N 300 Q. No. 93 Normality of a mixture of 30 ml of 1 N H SO 4 and 0 ml of 4 N H SO 4 is 1.0 N 1.1 N.0 N. N Explanation Total m. eq= 30 1 + 0 4 = 30 + 80 = 110 Total volume = 50 ml 110 N= =.N 50 Q. No. 94 Normality of solution obtained by mixing 10 ml of 1 N HCl, 0 ml of N H SO 4 and 30 ml of 3 N HNO 3 is 1.11 N. N.33 N 3.33 N Explanation 10 1+0 +30 3 N net = 60 =.33 N