May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
(m / 32) 1Moles fraction of O 2 ==(m / 32) + (m /16) 3According to Dalton's law of partial pressure,P ´O 2= P × mole fraction of O 2∴ Fraction of total pressure exerted by O 2 ,P´O= 21= Mole fraction of O2 =P3(b) Given that,Mass of O 2 = Mass of CH 4 = 32 g32No. of moles of O 2 = = 1 3232No. of moles of CH 4 = = 2 16According to Dalton's law of partial pressure,P ´O 2= P × mole fraction of O 2nRT= × Mole fraction of O2V3 × 0.0821×300 1=× = 24.63 atm1 3Similarly,P ´CH 4= P × mole fraction of CH 4nRT= × Mole fraction of CH4V3 × 0.0821×300 2=× = 49.26 atm1 3Total pressure, P = 24.63 + 49.26 = 73.89 atm4. A solution contains Na 2 CO 3 and NaHCO 3 .10 ml ofthis requires 2.0 ml of 0.1 M H 2 SO 4 for neutralizationusing phenolphthalein as indicator. Methyl orange isthen added when a further 2.5 ml of 0.2 M H 2 SO 4was required. Calculate the strength of Na 2 CO 3 andNaHCO 3 in solution.[IIT-1978]Sol. Step 1.Molecular massEquivalent mass of Na 2 CO 3 =2106= = 532mMeq. of Na 2 CO 3 in solution = 1× 1000 53Step 2.Molecular massEquivalent mass of NaHCO 3 =1= 84mMeq. of NaHCO 3 in solution = 2× 1000 84Step 3.Meq. of H 2 SO 4 used with phenolphthalein= Valency factor × Molarity × Volume (ml)= 2 × 0.1 × 2.0 = 0.42Na 2 CO 3 + H 2 SO 4 → 2NaHCO 3 + Na 2 SO 4Meq. of H 2 SO 4 used with phenolphthalein= 21 Meq. of Na2 CO 3 ∴ 21 Meq. of Na2 CO 3 = 0.4Step 4.Meq. of H 2 SO 4 used with methyl orange= Valency factor × molarity × volume(ml)= 2 × 0.2 × 2.5 = 1Meq. of H 2 SO 4 used with methyl orange= Meq. of NaHCO 3 + 21 Meq. of Na2 CO 3∴ Meq. of NaHCO 3 + 21 Meq. of Na2 CO 3 = 1∴ Meq. of NaHCO 3 = 1 – 0.4 = 0.6and Meq. of Na 2 CO 3 = 2 × 0.4 = 0.8Step 5.m 1 0 .8×53× 1000 = 0.8 or m 1 = = 0.04245310000 .0424×1000∴ Strength of Na 2 CO 3 solution =10= 4.24 g L –1Step 6.m 2× 1000 = 0.6 or m 2 =84∴ Strength of NaHCO 3 solution == 5.04 g L –10 .6×84= 0.050410000 .0504×1000105. Using the data given below, calculate the bondenthalpy of C–C and C–H bonds.∆ C Hº(ethane) = –1556.5 kJ mol –1∆ C Hº (propane) = –2117.5 kJ mol –1C(graphite) → C(g); ∆H = 719.7 kJ mol –1Bond enthalpy of H–H = 435.1 kJ mol –1∆ f Hº(H 2 O, 1) = –284.5 kJ mol –1∆ f Hº(CO 2 , g) = –393.3 kJ mol –1 [IIT-1990]Sol. From the enthalpy of combustion of ethane andpropane, we writeXtraEdge for IIT-JEE 36 MAY 2011
(1) C 2 H 6 (g) + 27O2 (g) → 2CO 2 (g) + 3H 2 O(1) :∆ C H = 3∆ f H(H 2 O, 1) + 2∆ f H(CO 2 , g) – ∆ f H(C 2 H 6 , g)Thus,∆ f H(C 2 H 6 ,g) = – ∆ C H + 3∆ f H(H 2 O, 1)+ 2∆ f H(CO 2 , g)= (1556.5 – 3 × 284.5 – 2 × 393.3) kJ mol –1= – 83.6 kJ mol –1(2) C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(1)∆ C H = 3∆ f H(CO 2 , g)+ 4∆ f H(H 2 O), 1) – ∆ f H(C 3 H 8 , g)Thus∆ f H(C 3 H 8 , g) = –∆ C H + 3∆ f H(CO 2 , g) + 4∆ f H(H 2 O, 1)= (2217.5 – 3 × 393.5 – 4 × 284.5) kJ mol –1= –101.0 kJ mol –1To calculate the ε C–H and ε C–C , we carry out thefollowing manipulations.(i) 2C(graphite) + 3H 2 (g) → C 2 H 6 (g)∆H = – 83.6 kJ mol –12C(g) → 2C (graphite)∆H = –2 × 719.7 kJ mol –16H(g) → 3H 2 (g)∆H = –3 × 435.1 kJ mol –1Add2C(g) + 6H(g) → C 2 H 6 (g)∆H (i) = (–83.6 – 2 × 719.7 – 3 × 435.1) kJ mol –1= – 2828.3 kJ mol –1(ii) 3C(graphite) + 4H 2 (g) → C 3 H 8 (g)∆H = –101.0 kJ mol –13C(g) → 3C (graphite)∆H = –3 × 719.7 kJ mol –18H(g) → 4H 2 (g)∆H = – 4 × 435.1 kJ mol –1Add3C(g) + 8H(g) → C 3 H 8 (g)∆H (ii) = (– 101 – 3 × 719.7 – 4 × 435.1) kJ mol –1= – 4000.5 kJ mol –1Now,∆H (i) = ε C–C – 6ε C–H= –2828.3 kJ mol –1∆H (ii) = –2ε C–C – 8ε C–H= –4000.5 kJ mol –1Solving for ε C–C and ε C–H , we getε C–H = 414.0 kJ mol –1and ε C–H = 344.3 kJ mol –1GalenaGalena is the natural mineral form of lead sulfide. Itis the most important lead ore mineral.Galena is one of the most abundant and widelydistributed sulfide minerals. It crystallizes in thecubic crystal system often showing octahedralforms. It is often associated with the mineralssphalerite, calcite and fluorite.Galena deposits often contain significant amountsof silver as included silver sulfide mineral phases oras limited solid solution within the galena structure.These argentiferous galenas have long been themost important ore of silver in mining. In additionzinc, cadmium, antimony, arsenic and bismuth alsooccur in variable amounts in lead ores. Seleniumsubstitutes for sulfur in the structure constituting asolid solution series. The lead telluride mineralaltaite has the same crystal structure as galena.Within the weathering or oxidation zone galenaalters to anglesite (lead sulfate) or cerussite (leadcarbonate). Galena exposed to acid mine drainagecan be oxidized to anglesite by naturally occurringbacteria and archaea, in a process similar tobioleaching [3]Galena uses :One of the earliest uses of galena was as kohl,which in Ancient Egypt, was applied around theeyes to reduce the glare of the desert sun and torepel flies, which were a potential source ofdisease.[4]Galena is a semiconductor with a small bandgap ofabout 0.4 eV which found use in early wirelesscommunication systems. For example, it was usedas the crystal in crystal radio sets, in which it wasused as a point-contact diode to detect the radiosignals. The galena crystal was used with a safetypin or similar sharp wire, which was known as a"cat's whisker". Making such wireless sets was apopular home hobby in the North of England duringthe 1930s. Derbyshire was one of the main areaswhere Galena was mined. Scientists that werelinked to this application are Karl Ferdinand Braunand Sir Jagdish Bose. In modern wirelesscommunication systems, galena detectors have beenreplaced by more reliable semiconductor devices,though silicon point-contact microwave detectorsstill exist in the market.XtraEdge for IIT-JEE 37 MAY 2011
- Page 3 and 4: Volume - 6 Issue - 11May, 2011 (Mon
- Page 5 and 6: Volume-6 Issue-11May, 2011 (Monthly
- Page 7 and 8: writer, broadcaster and member of t
- Page 9 and 10: KNOW IIT-JEEBy Previous Exam Questi
- Page 11 and 12: 60 cmK 1 K 2VB5. A solid sphere of
- Page 14 and 15: ∴ cos A ==cos B ==cos C ==b22+ c
- Page 16 and 17: Physics Challenging ProblemsSet # 1
- Page 18 and 19: PHYSICSStudents'ForumExpert’s Sol
- Page 20 and 21: Q 1 = 1200 JEnginel = 21 cma = 6 cm
- Page 22 and 23: PHYSICS FUNDAMENTAL FOR IIT-JEEElec
- Page 24 and 25: the indivual forces. It's often hel
- Page 26 and 27: Hence, E x = -∴∂V∂ xE = - ayi
- Page 28 and 29: For (x - a) 2 + (y - b) 2 = r 2 , t
- Page 30 and 31: (If the velocities are not along th
- Page 32 and 33: KEY CONCEPTPhysicalChemistryFundame
- Page 34 and 35: ⎛ n ⎞ ⎛ n ⎞Correction term
- Page 36 and 37: Kinetic Isotope Effects :The kineti
- Page 40 and 41: `tà{xÅtà|vtÄ V{tÄÄxÇzxá1 Se
- Page 42 and 43: Students' ForumExpert’s Solution
- Page 44 and 45: MATHSCOMPLEX NUMBERMathematics Fund
- Page 46 and 47: MATHSMATRICES &DETERMINANTSMathemat
- Page 48 and 49: aBased on New PatternIIT-JEE 2012Xt
- Page 50 and 51: dxO12. The time of crossing the riv
- Page 52 and 53: 6. Choose the INCORRECT statement f
- Page 54 and 55: 7. If the equation 3x 4 - 16x 3 + 3
- Page 56 and 57: Based on New PatternIIT-JEE 2013Xtr
- Page 58 and 59: 17. A stone is projected from level
- Page 60 and 61: 16. Half a mole of photon is used t
- Page 62 and 63: 17. Ifsin 3θcos 2θ2-147137= 0Then
- Page 64 and 65: IIT-JEE 2011PAPER-I (PAPER & SOLUTI
- Page 66 and 67: 9. Amongst the given options, the c
- Page 68 and 69: 22. The total number of alkenes pos
- Page 70 and 71: 1 2 × 8∆U = × [V - 0]22 2 + 81
- Page 72 and 73: shown in the figure. We use the sig
- Page 74 and 75: 40. A block is moving on an incline
- Page 76 and 77: MATHEMATICSSECTION - ISingle Correc
- Page 78 and 79: Ans.Sol.[A,D]r = xiˆ + yj ˆ + zk
- Page 80 and 81: ⎡323 21× ⎤=⎡32 1⎤1 C2C21 C
- Page 82 and 83: XtraEdge for IIT-JEE 80 MAY 2011
- Page 84 and 85: Ans. [A]Sol. ∆T = k f × m × i
- Page 86 and 87: SECTION - IVMatrix match TypeThis s
(1) C 2 H 6 (g) + 27O2 (g) → 2CO 2 (g) + 3H 2 O(1) :∆ C H = 3∆ f H(H 2 O, 1) + 2∆ f H(CO 2 , g) – ∆ f H(C 2 H 6 , g)Thus,∆ f H(C 2 H 6 ,g) = – ∆ C H + 3∆ f H(H 2 O, 1)+ 2∆ f H(CO 2 , g)= (1556.5 – 3 × 284.5 – 2 × 393.3) kJ mol –1= – 83.6 kJ mol –1(2) C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(1)∆ C H = 3∆ f H(CO 2 , g)+ 4∆ f H(H 2 O), 1) – ∆ f H(C 3 H 8 , g)Thus∆ f H(C 3 H 8 , g) = –∆ C H + 3∆ f H(CO 2 , g) + 4∆ f H(H 2 O, 1)= (2217.5 – 3 × 393.5 – 4 × 284.5) kJ mol –1= –101.0 kJ mol –1To calculate the ε C–H and ε C–C , we carry out thefollowing manipulations.(i) 2C(graphite) + 3H 2 (g) → C 2 H 6 (g)∆H = – 83.6 kJ mol –12C(g) → 2C (graphite)∆H = –2 × 719.7 kJ mol –16H(g) → 3H 2 (g)∆H = –3 × 435.1 kJ mol –1Add2C(g) + 6H(g) → C 2 H 6 (g)∆H (i) = (–83.6 – 2 × 719.7 – 3 × 435.1) kJ mol –1= – 2828.3 kJ mol –1(ii) 3C(graphite) + 4H 2 (g) → C 3 H 8 (g)∆H = –101.0 kJ mol –13C(g) → 3C (graphite)∆H = –3 × 719.7 kJ mol –18H(g) → 4H 2 (g)∆H = – 4 × 435.1 kJ mol –1Add3C(g) + 8H(g) → C 3 H 8 (g)∆H (ii) = (– 101 – 3 × 719.7 – 4 × 435.1) kJ mol –1= – 4000.5 kJ mol –1Now,∆H (i) = ε C–C – 6ε C–H= –2828.3 kJ mol –1∆H (ii) = –2ε C–C – 8ε C–H= –4000.5 kJ mol –1Solving for ε C–C and ε C–H , we getε C–H = 414.0 kJ mol –1and ε C–H = 344.3 kJ mol –1GalenaGalena is the natural mineral form of lead sulfide. Itis the most important lead ore mineral.Galena is one of the most abundant and widelydistributed sulfide minerals. It crystallizes in thecubic crystal system often showing octahedralforms. It is often associated with the mineralssphalerite, calcite and fluorite.Galena deposits often contain significant amountsof silver as included silver sulfide mineral phases oras limited solid solution within the galena structure.These argentiferous galenas have long been themost important ore of silver in mining. In additionzinc, cadmium, antimony, arsenic and bismuth alsooccur in variable amounts in lead ores. Seleniumsubstitutes for sulfur in the structure constituting asolid solution series. The lead telluride mineralaltaite has the same crystal structure as galena.Within the weathering or oxidation zone galenaalters to anglesite (lead sulfate) or cerussite (leadcarbonate). Galena exposed to acid mine drainagecan be oxidized to anglesite by naturally occurringbacteria and archaea, in a process similar tobioleaching [3]Galena uses :One of the earliest uses of galena was as kohl,which in Ancient Egypt, was applied around theeyes to reduce the glare of the desert sun and torepel flies, which were a potential source ofdisease.[4]Galena is a semiconductor with a small bandgap ofabout 0.4 eV which found use in early wirelesscommunication systems. For example, it was usedas the crystal in crystal radio sets, in which it wasused as a point-contact diode to detect the radiosignals. The galena crystal was used with a safetypin or similar sharp wire, which was known as a"cat's whisker". Making such wireless sets was apopular home hobby in the North of England duringthe 1930s. Derbyshire was one of the main areaswhere Galena was mined. Scientists that werelinked to this application are Karl Ferdinand Braunand Sir Jagdish Bose. In modern wirelesscommunication systems, galena detectors have beenreplaced by more reliable semiconductor devices,though silicon point-contact microwave detectorsstill exist in the market.XtraEdge for IIT-JEE 37 MAY <strong>2011</strong>