DPP-13 to 14 With Answer

FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 13 Class : XIII Course : DPP No.1 Max. Time : 40 Total Time : 40 min. Single choice Objective ('–1' negative marking) Q.1 to Q.5 (3 marks 3 min.) [15, 15] Subjective Questions ('–1' negative marking) Q.6 to Q.8 (4 marks 4 min.) [12, 12] REVISION QUESTIONS Single choice Objective ('–1' negative marking) Q.9 (3 marks 3 min.) [3, 3] Subjective Questions ('–1' negative marking) Q.10 (4 marks 4 min.) [4, 4] Comprehension ('–1' negative marking) Q.11 to Q.12 (3 marks 3 min.) [6, 6] 1. If numerical value of mass and velocity are equal then de-Broglie wavelength in terms of K.E. is : (1*) mh 2K.E. h (2) 2K.E. (3) both are correct (4) none is correct. h Sol.  = p h h = mv = v 2 2KE  but v2 = m therefore  = hm 2KE 2. Determine the de-Broglie wavelength associated with an electron in the 3rd Bohr's orbit of He+ ion? (1) 10 Å (2) 2 A (3*) 5 Å (4) 1 Å Sol. n = 2r 2r so  = 3 = 2 × (53 pm) × 9 5Å 3 2 3. If the light of wavelength 5.77 × 10–10 cm is used to detect an electron then the uncertainty in velocity (approximately) will be (h = 6.6 × 10–34 Js, me = 9.1 × 10–31 kg). (1) 106 m/s (2) 106 m/s (3*) 107 m/s (4) None of these Sol. V = h 4mx = 6.6 10 34 31 12 = 107 m/s. 4  3.14  9.110  5.77  10 4. Which of the following statement is true : . Higher is the mass of the particle, error in measurement of velocity reduces. . For an electron if uncertainty in position tends to zero then uncertainty in momentum is extremely small. . If 1 > 2 are the two different wavelength used to detect the position of electron and uncertainty in velocity be v1 and v2 respectively then v2 > v1 (1) only  (2)  and  (3) , &  (4*)  and  5. When ammonia is passed over heated copper oxide, the metallic copper is obtained. The reaction shows that ammonia is - (1) A dehydrating agent (2) An oxidising agent (3*) A reducing agent (4) A nitrating agent (–3) Sol. 3CuO + 2NH3 (R.A.)  3Cu + N2 + 3H2O. NH3 oxidized in N2 . So NH3 is reducing agent. 6. Identify the oxidant and the reductant in the following reactions : (1) Cu + HNO3 (dil)  Cu (NO3)2 + H2O + NO (2) Na2HAsO3 + KBrO3 + HCl  NaCl + KBr + H3AsO4 (0) ( 5) 2 2 Ans. (1) Cu + HNO3 (dil)  Cu (NO3)2 + H2O + NO . (0) 2 Cu (reductant)  Cu (NO ) (oxidation half). 5 2 HNO3 (oxidant)  NO (reduction half). 3 (2) Na2 HAsO3 + 3 5 K BrO3 + HCl  5 –1 NaCl + K Br 5 + H3 AsO4 Na2 HAsO3 (reductant)  H3 AsO4 (oxidation half). 5 KBrO3 (oxidant)  –1 K Br . 7. Balance the following redox equations and identify the valency factor (n-factor) for different compunds or molecule involved in the reactions at reactant and product side. (i) Al  [Al(OH) ]– + H (basic) Ans. 2OH– + 6H O + 2Al  3H + 2(Al(OH) ]– . 2 2 4 Sol. Mass Balance and Charge Balance : Oxidation Half : Al + 4OH–  Reduction Half : [Al(OH) ]– + 3e– . 2H O + 2e–  H + 2OH–. 2 2 Total loss electrons = total gain electrons. 2Al + 2OH– + 6H O  2[Al(OH) ]– + 3H . 2 4 2 Note : If H2 or O2 formed in reaction. It is means H2 or O2 produced from H2O molecule. (ii) Cu P + Cr O 2–  Cu2+ + H PO + Cr3+ + H O (acidic) 3 2 7 3 4 2 Cu P + Cr O 2–  Cu2+ + H PO + Cr3+ + H O (vEyh;) 3 2 7 3 4 2 Ans. 6Cu P + 11Cr O –2 + 124H+  18Cu+2 + 6H PO + 22Cr+3 + 53H O. 3 2 7 3 4 2 Sol. Mass Balance and Charge Balance : Oxidation Half : Cu P + 4H O  3Cu2+ + H PO + 5H+ + 11e–. 3 2 3 4 Reduction Half : Cr O 2– + 14H+ + 6e–  2Cr3+ + 7H O. 2 7 2 Total loss electrons = total gain electrons. 6Cu P + 11Cr O 2– + 124H+  18Cu2+ + 6H PO + 22Cr3+ + 53H O. 8. The reaction Cl (g) + S O 2-  SO 2- + Cl- is to be carried out in basic medium. Starting with 0.15 mol 2 2 3 4 of Cl , 0.01 mol S O 2- and 0.3 mol of OH-, how many moles of OH- will be left in solution after the 2 2 3 reaction is complete. Assume no other reaction occurs. Ans. 0.2 moles OH 2Cl + S O 2– + 10 OH–  2SO 2– + 4Cl– + 5H O S O 2– is L.R. so 0.2 moles of OH– will remain REVISION QUESTIONS 9. When a 12.0 g mixture of carbon and sulphur is burnt in air, then a mixture of CO2 and SO2 is produced, in which the number of moles of SO2 is half that of CO2. The mass of the carbon in the mixture is : (At. wt. S = 32) (1) 4.08 g (2*) 5.14 g (3) 8.74 g (4) 1.54 g Sol. (C + S)  CO + SO nSO2  nCO2 2 Let wt. of C = x So, wt. of S = 12 – x 12  x  1  x  then 32    x = 5.14 g. 2  12  10. How many grams of sodium dichromate, (Na2Cr2O7), should be added to a 50.0 mL volumetric flask to prepare 0.025 M Na2Cr2O7 when the flask is filled to the mark with water ? Ans. 0.3375 g. Mole Sol. Molarity = 0.025 = 50 50  0.025 × 1000  Mole = 1000 50  0.025 wt. of Na2Cr2O7 = 1000 × 270 = 0.3375 gm. Comprehension (Moderate) (Q.11 to Q.12) Molarity(mol/L) Molality(mol/Kg) Density (g/ml) Gram molecular mass of solute Solution-1 a - d1 P Solution-2 - b d2 Q Solution-3 1 - 1.060 60 11. What is molality of solution 1 : 1000a (1*) 1000 d1 aP 1000 d1 (2) 1000 aP a (3) 1000 x  ap (4)Noneofthese buesalsdksbZugha Sol. For solution 1  'a' moles of solute are present in 1000 ml of solution. wt. of solution = 1000 × d g  wt. of solute = aP g  a  1000  So, Molality = 1000  d1  aP   12. What is the molarity of solution 2 : b x d2 1000 bQ (2*) b x 1000 x d2 1000  bQ 1000 x bQ (3) 1000 bd2 (4) NoneofthesebuesalsdksbZugha Sol. For solution 2  'b' moles of solute are present in 1000 g of solvent. wt. of solution = 1000 + bQ vol. of solution = 1000  bQ d2  Molality = b 1000 1000  bQ d2 b 1000  d2 = 1000  bQ FACULTY COPY DAILY PRACTICE PROBLEMS (DPP) Subject : Physical/Inorg.Chemistry Date : DPP No. 14 Class : XIII Course DPP No.2 Max. Marks : 39 Total Time : 39 min. Single choice Objective ('–1' negative marking) Q.1 to Q.8 (3 marks 3 min.) [24, 24] Subjective Questions ('–1' negative marking) Q.9 to Q.10 (4 marks 4 min.) [8, 8] REVISION QUESTIONS Single choice Objective ('–1' negative marking) Q.11 (3 marks 3 min.) [3, 3] Subjective Questions ('–1' negative marking) Q.12 (4 marks 4 min.) [4, 4] 1. The uncertainty in position and velocity of the particle are 0.1 nm and 5.27×10–24 ms–1 respectively then the mass of the particle is : (h = 6.625 × 10–34Js) (1) 200 g (2) 300 g (3*) 100 g (4) 1000 g Sol. m = h 4xv = 6.6 1034 4  3.14 1010  5.27 1024 ~ 100 g. 2. Manganese achieves its maximum oxidation state in its compounds (1) MnO2 (2) Mn3O4 (3*) KMnO4 (4) K2MnO4 Sol. (7) KMnO4 (  6) > K2MnO4 ( 4) > MnO2 8/3 > Mn3O4 3. The molecular weight of the compounds (a) Na2SO4, (b) Na3PO4. 12H2O and (c) Ca3(PO4)2 respectively are X, Y and Z. the correct set of their equivalent weights will be - (1*) (a) X 2 (b) X Y 3 (c) Z Z 6 (2) (a) X (b) Y Z 3 (c) 3 (3) (a) 2 (b) Y (c) 3 (4) (a) X (b) Y (c) Z Sol. n = Total charge of cation or total charge of anion. If reaction is given (with a salt not a redox reaction) — Then V.F. (n) = no. of charges (+ ve/ – ve)being displaced or taking part in reaction. (a) Na SO X Y E = . (b) Na PO .12H O E = 2 4 Na2SO4 2 3 4 2 Na3PO4 3 (c) Ca (PO ) Z E = . 3 4 2 Ca3 (PO4 )2 6 4. m grams of a mixture of Na2CO3 and NaHCO3 yields one mole of CO2 on treatment with HCl. After this reaction the solution is evaporated and yields 1.4 mole of NaCl. What is the value of m? (1) 95 gm (2) 101 gm (3) 97.2 gm (4*) 92.4 gm Sol. Let moles of Na2CO3 & NaHCO3 are x and y respectively Na2CO3 + 2HCl NaHCO3 + HCl x + y = 1 2x + y = 1.4  2NaCl + H O + CO  NaCl + H O + CO 5. A vessel of 120 ml capacity contains a certain mass of a gas at 20ºC and 750 mm pressure. The gas was transferred to a vessel, whose volume is 180 ml, then the pressure of gas at 20ºC is : (1*) 500 mm (2) 250 mm (3) 1000 mm (4) None of these 6. 2.5 L of a sample of a gas at 27°C and 1 bar pressure is compressed to a volume of 500 mL keeping the temperature constant, the percentage increase in pressure is - (1) 100 % (2*) 400 % (3) 500% (4) 80% 7. If V0 is the volume of a given mass of gas at 273 K at a constant pressure, then according to Charles' law, the volume at 10ºC will be : 1 10 283 (1) 10 V0 (2) 273 (V0 + 10) (3) V0 + 273 (4*) 273 V0 8. For the four gases A, B, E and D the value of the excluded volume per mole is same. If the order of the critical temperature is TB > TD > TA > TE then the order of their liquefaction pressure at a temperature T (T < TE) will be : (1) PA < PB < PE < PD (2*) PB < PD < PA < PE (3) PE < PA < PD < PB (4) PD < PE < PA < PB 9. Calculate ratio of de-Broglie wavelength for a proton and -particle if their K.E. are same. Ans. p  2 .  1 Sol. Use  = 10. Balance the following redox equations and identify the valency factor (n-factor) for different compunds or molecule involved in the reactions at reactant and product side. (i) ClO3– + Fe2+ + H+  Cl– + Fe3+ + H2O Ans. 6H+ + ClO3– + 6Fe2+  Cl– + 6Fe3+ + 3H2O. Sol. Mass Balance and Charge Balance : Oxidation Half : Fe2+  Fe3+ + 1e–. Reduction Half : ClO – + 6H+ + 6e–  Cl– + 3H O. 3 2 Total loss electrons = total gain electrons. 6Fe2+ + 6H+  6Fe3+ + Cl–+ 3H O. (ii) N2O4 + BrO3–  NO3– + Br– Ans. 3N2O4 + BrO3– + 3H2O  6NO3– + Br– + 6H+. Sol. Mass Balance and Charge Balance : Oxidation Half : N O + 2H O  2NO – + 4H+ + 2e–. 2 4 2 3 Reduction Half : BrO – + 6H+ + 6e–  Br– + 3H O. 3 2 Total loss electrons = total gain electrons. 3N O + 3H O + BrO –  6NO – + 6H+ + Br–. REVISION QUESTIONS 11. A mixture of N2 and H2 is caused to react in a closed container to form NH3 . The reaction ceases before either reactant has been totally consumed. At this stage, 2.0 moles each of N2, H2 and NH3 are present. The moles of N2 and H2 present originally were respectively : (1) 4 and 4 moles (2*) 3 and 5 moles (3) 3 and 4 moles (4) 4 and 5 moles Sol. N2 + 3H2  2NH . a b 0 (a–x) (b–3x) 2x Given, 2x = 2, So x = 1. a – x = 2, So a = 2 + 1 = 3. b – 3x = 2, So b = 2 + 3 × 1 = 5. 12. 10 cc of H2O2 solution when reacted with K solution produced 1.27 g of iodine with KOH. Calculate the percentage strength of H2O2 .(% w/v)  = 127] Ans. 1.7%. Sol. H O + 2K  2 + 2KOH 1.27 Mole of  = 254 = 0.05 = Mole of H2O2 wt. of H2O2 = 0.05 × 34 = 0.17 g. 0.17 So, % w/v = 10 × 100 = 1.7%.