Solution - Career Point

39.[B] Solution is decinormal, that is N/10 & x factoris 1, so conc. = 0.1M[H + ] = c.α= 0.1 ×1.3/100 = 13×10 –4pH = –log [13×10 –4 ] = 2.8940.[A] The no. of atoms in fcc lattice = 4a = 400 pm = 4×10 –10 m= 4×10 –8 cmn × M 4×60d = =323–8 3No × a 6×10 × (4×10 )d = 6.23 g/cm 3 .41.[B] For any Value of l possible values of m arem = –l to + ll = 2, m = –2, –1, 0 +1, +2So option is (B)42.[B] C : O : H6gm : 3.01 × 10 23 atoms : 2 moleRatio ½ : ½ : 2of mole1 : 1 : 4COH 4 or CH 4 O43.[B]44.[B]45. [A]312NaNO 2 /HCl gives HNO 2 which gives differentproducts with Pri. and Sec. amines.CH 3+ Cl 2⎯⎯→h νCH 2 Cl +HCl46.[D] 2 CuSO 4 + 2 KCN ⎯→ Cu(CN) 2 + K 2 SO 42Cu(CN) 2⎯⎯→ 2CuCN ↓ + (CN) 2 ↑3 KCN + CuCN ⎯⎯→ K 3 [Cu(CN) 4 ]47. [B] Blue print process occurs with the help of IronCompound.48.[A] Effective nuclear charge increases thereforeionic radius follow the order.49.[B] xy 2 xy + yt = 0 : 600 0 0teq m : 600-P P PNow : 600 – P + P + P = 800P = 200 mm Hg(Pxy)(Py) 200×200Kp = =(Pxy ) 4002= 100mm50.[C] Cell reaction : Zn + Cu +2 ⎯→ Zn +2 + CuCell emf:Hg0.059 ⎛E cell = E° cell – log ⎜Znn⎝ CuSo doubling the conc. of ions.E cell remains same.+ 2+ 251.[A] pK a = –log K a = –log (1.8×10 –5 ) = 4.744712 mol[CH 3 COOH] = = 0.460×0.5 L16.4 mol[CH 3 COONa] = = 0.482× 0.5 L⎡ salt ⎤Now, pH = pK a + log ⎢ ⎥⎣acid⎦⎞⎟⎠⎡0.4⎤= 4.7447 + log ⎢ ⎥⎣0.4= 4.7447 ⎦Increase = 152.[B] FeC 2 O 4 ⎯→ Fe +3 + CO 2+2 +3 +3 +4Increase = 1Total Increase in O.N. = 3So valence factor of FeC 2 O 4 = 3KMnO 4 ⎯→Mn +2+7 +2 v.f. (KMnO 4 ) = 5gm E KMnO 4 = gm E of FeC 2 O 4Mole × v.f. = Mole × v.f.1× 5 = x × 3 ⇒ x = 5/353.[A] ∆G = ∆H – T ∆S∆G = Θ , ∆G < 0 (Spontaneous process)∆G = ∆H – T ∆S= ∆E + P ∆V – T ∆S(∆G) E,V = 0 + 0 – T ∆S(∆S) E,V = ⊕ ⇒ ∆G = Θ (Spontaneous process)54.[B]CH| 3H3C – C – CH – CH|3OH| CH3CH 3 |H++⎯⎯→CH3– C – C H – CH–H O32CH| 32°Carbocation1, 2-Methyl shift+ +H C – C C– CH –H3H C – C – CH – CHCH| =CH| 3 ←⎯⎯3CH| CH| 33 33 33°Carbocation55.[C] Factual Q.(Morestable)XtraEdge for IIT-JEE 105APRIL 2010