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Brain Storming Comprehensions (Chemistry)

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   Aakash Educational Services Ltd. - Regd. Office:  Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 Linked Comprehension Type C1.1. Answer (3)Concentration of H 2 CO 3  in rain water = M1006.7 102053.3 56 − ×=× When NaOH is added the following reaction takes placeH 2 CO 3  + NaOH NaHCO 3  + H 2 Oinitial conc.7.06 × 10  –5  Mafter reaction conc.(7.06 × 10  –5  – 3 × 10  –5 ) M 3 × 10  –5 M = 4.06 × 10  –5 Resulting solution will contain NaHCO 3  and H 2 CO 3  hence it will act as an acidic buffer pH = pK a  + ]acid[]salt[log = )1006.4( )103( log35.6 55 −− ××+ = 6.35 – 0.13 = 6.222. Answer (3)The equilibria that co-exist areCd(OH 2 )(s) x2)yx( 2 OH2Cd  −−+ + …(i) zya232yx2 OSCd −−−−+ +   zy32 )OS(Cd − …(ii) zya232zy32 OS)OS(Cd −−−− +   z2232 ])OS(Cd[  − …(iii)Let x be the moles of Cd(OH) 2  that dissolvesK sp  = (x – y)(2x) 2 4.5 × 10  –15  = (5 × 10  –4  – y)(10  –3 ) 2 y = 4.99 × 10  –4   ~  5 × 10  –4 [Cd 2+ ] left to maintain the equilibrium isM105.4 )10(105.4 ]OH[K 923152sp  −−−−  ×=×= Using (ii) and (iii) equation Brain StormingComprehensions UNIT   5  206 Brain Storming Comprehensions Success Magnet -Solutions (Part-I) Aakash Educational Services Ltd. - Regd. Office:  Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 K 1 K 2  = 223222232 ]OS][Cd[ ])OS(Cd[ −+− Let the conc. of thiosulphate at equilibrium be b mole/litre8.3 × 10 3  × 2.5 × 10 2  = 29 b)105.4( z − ×  As the equilibrium constant for reaction of complexation of Cd 2+  with thiosulphate is high therefore it canbe assumed that z ~  5 × 10  –4 b 2  = 0535.0105.410075.2 105 964 =×××× −− b = 0.231 MSince value of y and z are very small ( i.e. , 5 × 10  –4 ) therefore amount of Na 2 S 2 O 3  needed to dissolve5 × 10  –4  mole of Cd(OH) 2  is 0.231 moles.3. Answer (1)pH = pK a  + log ]acid[]salt[  = 4.74 + log5.025.0 = 4.438pOH = 9.562[OH  –  ] = 2.74 × 10  –10 Ionic product of Mg(OH) 2  = [Mg 2+ ][OH  –  ] 2  = (0.1)(2.74 × 10  –10 ) 2  = 7.5 × 10  –21 Ionic product of Al(OH) 3  = (0.1)(2.74 × 10  –10 ) 3  = 2.05 × 10  –30 Ionic product of Fe(OH) 3  = 0.1(2.74 × 10  –10 ) 3  = 2.05 × 10  –30 In case of only Mg(OH) 2  ionic product < K sp C2.1. Answer (3)Electronic configuration of Be +  is 1 s 2 , 2 s 1 Energy of last electron in Be +  is given byE = 22.eff  nZ6.13 −  –17.98 = 22.eff  nZ6.13 − ∴ Z eff.  = 2.3Extent of shielding by inner electrons = 4 – 2.3 = 1.7 unitfor Li, Z eff.  = 3 – 1.7 = 1.3E = eV223.13.1 6.13 ×××−  = –5.746 eVIonisation energy of lithium = 5.746 eV.  207 Success Magnet -Solutions (Part-I) Brain Storming Comprehensions  Aakash Educational Services Ltd. - Regd. Office:  Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 2. Answer (3)|F| = 42 r kedr du = …(i) r mvr ke 242 = …(ii)and π= 2nhmvr …(iii)Equating (ii) and (iii)r = 212222 nmkhnm4ke =π Total energy = 21(potential energy) = 33162321232 mk6nkenmk6ker 6ke  −=⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ −=− Total energy ∝  n 6 Total energy ∝  m  –3 .3. Answer (3)Because of 6 different lines are observed in H-atom spectrum highest energy level is n = 4. Also theradiations responsible for photoelectric effect are due to transition formn = 4 to n = 1andn = 3 to n = 1 Δ E 1  = eV75.12 16116.13  =⎥⎦⎤⎢⎣⎡− Δ E 2  = eV1.12 9116.13  =⎥⎦⎤⎢⎣⎡− K 1  = Δ E 1  – φ , K 2  = Δ E 2  – φ 5 = φ−φ− 1.1275.12 φ  = 11.94 eVC3.1.Answer (2)In propane, number of 1ºH atoms are 6 and number of 2º H atoms are 2. Product (x) is formed byreplacement of 1ºH and product (y) is formed by replacement of 2ºH. So, the product ratio ⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛  yx  is 13.2.Answer (3)In case of Br, the abstraction of 3ºH is 1600 times in comparison to 1ºH. So, the % of x product is about99%.  208 Brain Storming Comprehensions Success Magnet -Solutions (Part-I) Aakash Educational Services Ltd. - Regd. Office:  Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456 3.Answer (1) CH – CH – CH + Cl 332 CH 3 h ν CH – CH – CH – Cl 32 CH 3 +CH – C – CH 33 CH 3 (II)Cl(I) Possibility of (I)= No. of 1ºH × reactivity of 1ºH= 9 × 1 = 9Possibility of (II)= No. of 3ºH × reactivity of 3ºH= 1 × 5 = 5% of (I) = %3.64100 149 =× % of (II) = %7.35100 145 =× Hence, major product is CH – CH – CH – Cl 32 CH 2 C4.1. Answer (2)2. Answer (2) As the electron emitted with zero velocity i.e.  with no kinetic energy so energy given will be equal tothreshold energy (work function).Hence, 0 hhc ν=λ or, 1141080 sec1028.4 107000103c  −−  ×=××=λ=ν 3. Answer (2)Only Li is not able to give photoelectric effect.C5.1. Answer (1)P 2 V = constantT 2 V  –1  = constant (by the ideal gas equation)T 2  = kVT 12  = k 5VT 1  = T5Internal energy of a gas is function of its temperature therefore II is correct.2. Answer (4)w = ∫∫  = dVkVnRdV·P …(i)T = kV 2 PV = nRTp = nRkV
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