Page 1 CBSE XII | Chemistry Sample Paper 3 - Solution CBSE Class XII Chemistry Sample Paper 3 - Solution Time: 3 Hrs Maximum Marks: 70 _________________________________________________________________________________________________________ Section A 1. A polymer formed by the polymerisation of two or more different monomers is called a copolymer; for example, Buna-S. 2. This is because amorphous solids have a tendency to flow. OR Frenkel defect is shown by ionic compounds having large difference in the size of ions so that the smaller ion dislocates from its normal site and occupies an interstitial site. In case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy interstitial sites. 3. Pyridinium chlorochromate 4. Yes, if one of the reactants is taken in excess, its concentration does not change considerably with time. So, a bimolecular reaction can be of the first order. 5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be used for drying ammonia. OR The increasing order of acidic character of haloacids is HF < HCl < HBr < HI Section B 6. (a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the number of moles of that component to the total number of moles of all the components of the mixture. Mole fraction is given by nA = Molesof A Total,molesof allthecomponentsof themixtur e (b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as Observedvalueof colligativeproperty i= Normalvalueof colligativeproperty OR Page 2 CBSE XII | Chemistry Sample Paper 3 - Solution CBSE Class XII Chemistry Sample Paper 3 - Solution Time: 3 Hrs Maximum Marks: 70 _________________________________________________________________________________________________________ Section A 1. A polymer formed by the polymerisation of two or more different monomers is called a copolymer; for example, Buna-S. 2. This is because amorphous solids have a tendency to flow. OR Frenkel defect is shown by ionic compounds having large difference in the size of ions so that the smaller ion dislocates from its normal site and occupies an interstitial site. In case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy interstitial sites. 3. Pyridinium chlorochromate 4. Yes, if one of the reactants is taken in excess, its concentration does not change considerably with time. So, a bimolecular reaction can be of the first order. 5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be used for drying ammonia. OR The increasing order of acidic character of haloacids is HF < HCl < HBr < HI Section B 6. (a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the number of moles of that component to the total number of moles of all the components of the mixture. Mole fraction is given by nA = Molesof A Total,molesof allthecomponentsof themixtur e (b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as Observedvalueof colligativeproperty i= Normalvalueof colligativeproperty OR CBSE XII | Chemistry Sample Paper 3 - Solution The relationship between mole fraction and vapour pressure of a component of an ideal solution in the liquid phase and vapour phase: 0 A A A 0 B B B AB 00 A A B B P =x .P P =x .P P =P +P = x .P +x .P 7. When Kc < 1, taking log, gives a negative value. For example: Thus, is negative if the equilibrium constant Kc < 1. 8. Given: Rate constant K = 200 s -1 The half-life of the first-order reaction is 3 1 2 0.693 0.693 t 3.46 10 sec. K 200 ? ? ? ? ? 9. (a) Cl2O: 2x – 2 = 0 2x = 2 x = + 1 (b) KBrO3: 1 + x - 6 = 0 x = 6 + 1 x = 7 cell c 0.059 For a cell, E = log K n ? cell 0.059 E = log 0.01 n = - 2 x 0.059/n ( negative value) ? cell E ? cell c c c c If E 0 0.059 Then 0 = log K n log K 0 K Antilog(0) K = 1 ? ? ? ? ? Page 3 CBSE XII | Chemistry Sample Paper 3 - Solution CBSE Class XII Chemistry Sample Paper 3 - Solution Time: 3 Hrs Maximum Marks: 70 _________________________________________________________________________________________________________ Section A 1. A polymer formed by the polymerisation of two or more different monomers is called a copolymer; for example, Buna-S. 2. This is because amorphous solids have a tendency to flow. OR Frenkel defect is shown by ionic compounds having large difference in the size of ions so that the smaller ion dislocates from its normal site and occupies an interstitial site. In case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy interstitial sites. 3. Pyridinium chlorochromate 4. Yes, if one of the reactants is taken in excess, its concentration does not change considerably with time. So, a bimolecular reaction can be of the first order. 5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be used for drying ammonia. OR The increasing order of acidic character of haloacids is HF < HCl < HBr < HI Section B 6. (a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the number of moles of that component to the total number of moles of all the components of the mixture. Mole fraction is given by nA = Molesof A Total,molesof allthecomponentsof themixtur e (b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as Observedvalueof colligativeproperty i= Normalvalueof colligativeproperty OR CBSE XII | Chemistry Sample Paper 3 - Solution The relationship between mole fraction and vapour pressure of a component of an ideal solution in the liquid phase and vapour phase: 0 A A A 0 B B B AB 00 A A B B P =x .P P =x .P P =P +P = x .P +x .P 7. When Kc < 1, taking log, gives a negative value. For example: Thus, is negative if the equilibrium constant Kc < 1. 8. Given: Rate constant K = 200 s -1 The half-life of the first-order reaction is 3 1 2 0.693 0.693 t 3.46 10 sec. K 200 ? ? ? ? ? 9. (a) Cl2O: 2x – 2 = 0 2x = 2 x = + 1 (b) KBrO3: 1 + x - 6 = 0 x = 6 + 1 x = 7 cell c 0.059 For a cell, E = log K n ? cell 0.059 E = log 0.01 n = - 2 x 0.059/n ( negative value) ? cell E ? cell c c c c If E 0 0.059 Then 0 = log K n log K 0 K Antilog(0) K = 1 ? ? ? ? ? CBSE XII | Chemistry Sample Paper 3 - Solution 10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of iodoform, while methanol will not. C2H5OH + 4I2 + 6NaOH CHI3 + 5NaI + 5H2O + HCOONa Yellow CH3OH + I2 + NaOH no yellow ppt. 11. (a) Methyl-2-aminobutanoate (b) N-methylbenzenamine OR (a) Ambident nucleophile (i) 2 5 2 5 C H Cl+KCN C H CN+KCl ? ? ? (ii) 2 5 2 5 C H Cl+AgCN C H -N C+AgCl ? ? ? ? From the above reactions, it is clear that CN - is an ambident nucleophile. (b) Hinsberg test: ? ? Page 4 CBSE XII | Chemistry Sample Paper 3 - Solution CBSE Class XII Chemistry Sample Paper 3 - Solution Time: 3 Hrs Maximum Marks: 70 _________________________________________________________________________________________________________ Section A 1. A polymer formed by the polymerisation of two or more different monomers is called a copolymer; for example, Buna-S. 2. This is because amorphous solids have a tendency to flow. OR Frenkel defect is shown by ionic compounds having large difference in the size of ions so that the smaller ion dislocates from its normal site and occupies an interstitial site. In case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy interstitial sites. 3. Pyridinium chlorochromate 4. Yes, if one of the reactants is taken in excess, its concentration does not change considerably with time. So, a bimolecular reaction can be of the first order. 5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be used for drying ammonia. OR The increasing order of acidic character of haloacids is HF < HCl < HBr < HI Section B 6. (a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the number of moles of that component to the total number of moles of all the components of the mixture. Mole fraction is given by nA = Molesof A Total,molesof allthecomponentsof themixtur e (b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as Observedvalueof colligativeproperty i= Normalvalueof colligativeproperty OR CBSE XII | Chemistry Sample Paper 3 - Solution The relationship between mole fraction and vapour pressure of a component of an ideal solution in the liquid phase and vapour phase: 0 A A A 0 B B B AB 00 A A B B P =x .P P =x .P P =P +P = x .P +x .P 7. When Kc < 1, taking log, gives a negative value. For example: Thus, is negative if the equilibrium constant Kc < 1. 8. Given: Rate constant K = 200 s -1 The half-life of the first-order reaction is 3 1 2 0.693 0.693 t 3.46 10 sec. K 200 ? ? ? ? ? 9. (a) Cl2O: 2x – 2 = 0 2x = 2 x = + 1 (b) KBrO3: 1 + x - 6 = 0 x = 6 + 1 x = 7 cell c 0.059 For a cell, E = log K n ? cell 0.059 E = log 0.01 n = - 2 x 0.059/n ( negative value) ? cell E ? cell c c c c If E 0 0.059 Then 0 = log K n log K 0 K Antilog(0) K = 1 ? ? ? ? ? CBSE XII | Chemistry Sample Paper 3 - Solution 10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of iodoform, while methanol will not. C2H5OH + 4I2 + 6NaOH CHI3 + 5NaI + 5H2O + HCOONa Yellow CH3OH + I2 + NaOH no yellow ppt. 11. (a) Methyl-2-aminobutanoate (b) N-methylbenzenamine OR (a) Ambident nucleophile (i) 2 5 2 5 C H Cl+KCN C H CN+KCl ? ? ? (ii) 2 5 2 5 C H Cl+AgCN C H -N C+AgCl ? ? ? ? From the above reactions, it is clear that CN - is an ambident nucleophile. (b) Hinsberg test: ? ? CBSE XII | Chemistry Sample Paper 3 - Solution 12. (a) Benzoyl peroxide acts as an initiator because it generates free radicals. (b) LDPE is low-density polyethene, whereas HDPE is high-density polyethene. LDPE: It is produced by free radical polymerisation at a high temperature of 350– 570 K and a high pressure of 1000–2000 atm. It is a branched chain polymer. HDPE: It is produced by polymerisation of ethene in the presence of Ziegler Natta catalyst at a temperature of 333–343 K and under a pressure of 6–7 atm. It is a linear polymer. Section C 13. Given: Cell edge (a) = 288 pm = 288 ×10 -10 cm Let’s first find the volume, Volume of unit cell = (288 ×10 -10 cm) 3 = 2.389×10 -23 cm 3 3 Mass Volume of 208 g of the element Density 208 7.2 28.89cm ? ? ? 23 23 3 Totalvolume Numberof unit cells Volumeof aunit cell 28.89 2.389 10 12.09 10 cm ? ? ? ? ? ?? For a b.c.c. structure, the number of atoms per unit cell = 2 Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells = 2 × 12.09 × 10 23 = 24.18 × 10 23 = 2.418×10 24 atoms Page 5 CBSE XII | Chemistry Sample Paper 3 - Solution CBSE Class XII Chemistry Sample Paper 3 - Solution Time: 3 Hrs Maximum Marks: 70 _________________________________________________________________________________________________________ Section A 1. A polymer formed by the polymerisation of two or more different monomers is called a copolymer; for example, Buna-S. 2. This is because amorphous solids have a tendency to flow. OR Frenkel defect is shown by ionic compounds having large difference in the size of ions so that the smaller ion dislocates from its normal site and occupies an interstitial site. In case of halides of alkali metals, ions of alkali metal halides are larger and cannot occupy interstitial sites. 3. Pyridinium chlorochromate 4. Yes, if one of the reactants is taken in excess, its concentration does not change considerably with time. So, a bimolecular reaction can be of the first order. 5. P4O10 reacts with moist ammonia forming ammonium phosphate. Thus, it cannot be used for drying ammonia. OR The increasing order of acidic character of haloacids is HF < HCl < HBr < HI Section B 6. (a) Mole fraction (x): Mole fraction of a component in a mixture is the ratio between the number of moles of that component to the total number of moles of all the components of the mixture. Mole fraction is given by nA = Molesof A Total,molesof allthecomponentsof themixtur e (b) van’t Hoff factor: van’t Hoff factor ‘I’ is a correction factor defined as Observedvalueof colligativeproperty i= Normalvalueof colligativeproperty OR CBSE XII | Chemistry Sample Paper 3 - Solution The relationship between mole fraction and vapour pressure of a component of an ideal solution in the liquid phase and vapour phase: 0 A A A 0 B B B AB 00 A A B B P =x .P P =x .P P =P +P = x .P +x .P 7. When Kc < 1, taking log, gives a negative value. For example: Thus, is negative if the equilibrium constant Kc < 1. 8. Given: Rate constant K = 200 s -1 The half-life of the first-order reaction is 3 1 2 0.693 0.693 t 3.46 10 sec. K 200 ? ? ? ? ? 9. (a) Cl2O: 2x – 2 = 0 2x = 2 x = + 1 (b) KBrO3: 1 + x - 6 = 0 x = 6 + 1 x = 7 cell c 0.059 For a cell, E = log K n ? cell 0.059 E = log 0.01 n = - 2 x 0.059/n ( negative value) ? cell E ? cell c c c c If E 0 0.059 Then 0 = log K n log K 0 K Antilog(0) K = 1 ? ? ? ? ? CBSE XII | Chemistry Sample Paper 3 - Solution 10. On adding I2 and NaOH to both ethanol and methanol, ethanol will give a yellow ppt. of iodoform, while methanol will not. C2H5OH + 4I2 + 6NaOH CHI3 + 5NaI + 5H2O + HCOONa Yellow CH3OH + I2 + NaOH no yellow ppt. 11. (a) Methyl-2-aminobutanoate (b) N-methylbenzenamine OR (a) Ambident nucleophile (i) 2 5 2 5 C H Cl+KCN C H CN+KCl ? ? ? (ii) 2 5 2 5 C H Cl+AgCN C H -N C+AgCl ? ? ? ? From the above reactions, it is clear that CN - is an ambident nucleophile. (b) Hinsberg test: ? ? CBSE XII | Chemistry Sample Paper 3 - Solution 12. (a) Benzoyl peroxide acts as an initiator because it generates free radicals. (b) LDPE is low-density polyethene, whereas HDPE is high-density polyethene. LDPE: It is produced by free radical polymerisation at a high temperature of 350– 570 K and a high pressure of 1000–2000 atm. It is a branched chain polymer. HDPE: It is produced by polymerisation of ethene in the presence of Ziegler Natta catalyst at a temperature of 333–343 K and under a pressure of 6–7 atm. It is a linear polymer. Section C 13. Given: Cell edge (a) = 288 pm = 288 ×10 -10 cm Let’s first find the volume, Volume of unit cell = (288 ×10 -10 cm) 3 = 2.389×10 -23 cm 3 3 Mass Volume of 208 g of the element Density 208 7.2 28.89cm ? ? ? 23 23 3 Totalvolume Numberof unit cells Volumeof aunit cell 28.89 2.389 10 12.09 10 cm ? ? ? ? ? ?? For a b.c.c. structure, the number of atoms per unit cell = 2 Number of atoms present in 208 g = No. of atoms per unit cell × No. of unit cells = 2 × 12.09 × 10 23 = 24.18 × 10 23 = 2.418×10 24 atoms CBSE XII | Chemistry Sample Paper 3 - Solution 14. Given: 1 1 1 2 1 1 1 0 1 1 2 2 1 0 1 1 2 2 1 0 11 0 1 1 0 1 1 0 1 1 0 1 1 1 W 90g W 10g M 72 12 96 180g mol P 32.8mmHg W 18g mol P P n W M P n n M W PP 92 18 P 180 90 P 92 18 1 P 180 90 P 92 18 1 P 180 90 P 92 18 1 P 180 90 P 92 18 1 32.8 180 90 P 0.36KPa ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ?? ?? ? ?? ? ?? ? ? OR 8 18 o oo s B B 1 CH 1 80 P P 80%of P 0.80P 100 W n mol 40 114g n 1mol...................(Molar mass 114gmol) 114gmol ? ? ? ? ? ? ? ? ? 8 18 B os B B o B B C H w n pp 40 Molesof octane x W p n n 40 1 ? ? ? ? ? ? ? ? B oo o B B B w p 0.80P 40 W P 40 1 w W 40 1 0.80 W W 40 40 1 ? ? ? ? ? ? ? ?Read More
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