September 2011 - Career Point

Given that K sp for Ag 2 C 2 O 4 = 1.29 × 10 –11 mol 3 l –3 at25ºCSo, [Ag + ] 2 [C 2 O 2– 4 ] = 1.29 × 10 –11or [Ag + ] 2 × 0.2324 = 1.29 × 10 –11Hence [Ag+] 2 1.29= × 10 –110.2324Then K sp for−11Ag 2 CO 3 = [Ag + ] 2 [CO 2– 1.29×103 ] =× 0.07160.2324= 3.794 × 10 –12 mol 3 l –310. The molar volume of liquid benzene(density = 0.877 g ml –1 ) increases by a factor of 2750as it vaporizes at 20ºC and that of liquid toluene(density = 0.867 g ml –1 ) increases by a factor of 7720at 20ºC. A solution of benzene and toluene at 20ºChas a vapour pressure of 46.0 torr. Find the molefraction of benzene in vapour above the solution.[IIT-1996]Sol. Given that,Density of benzene = 0.877 g ml –1Molecular mass of benzene (C 6 H 6 )= 6 × 12 + 6 × 1 = 7878∴ Molar volume of benzene in liquid form = ml 0.877=780.8771× 1000L = 244.58 LAnd molar volume of benzene in vapour phse=780.8772750× L = 244.58 L1000Density of toluene = 0.867 g ml –1Molecular mass of toluene (C 6 H 5 CH 3 )= 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92∴ Molar volume of toluene in liquid form92 92 1= ml = × L0.867 0. 867 1000And molar volume of toluene in vapour phase=920.8677720× L = 819.19 L1000Using the ideal gas equation,PV = nRTAt T = 20ºC = 293 KFor benzene, P =0P B =nRTV1 × 0.082×293=244.58= 0.098 atm= 74.48 torr (Q 1 atm = 760 torr)Similarly, for toluene,P =0P T =nRTV1 × 0.082×293== 0.029 atm819.19= 22.04 torr (Q 1 atm = 760 torr)According to Raoult's law,0P B = PBx B = 74.48 x B0P T = PTx T = 22.04 (1 – x B )0 0And P M = PBx B + PTx Tor 46.0 = 74.48 x B + 22.04 (1 – x B )Solving, x B = 0.457According to Dalton's law,'P B = P M x B (in vapour phase)or mole fraction of benzene in vapour form,'x B =PPBM74 .48×0.457== 0.7446.0MATHEMATICSt –te + e e t – e–t11. For any real t, x = , y = is a point22on the hyperbola x 2 – y 2 = 1. Find the area boundedby this hyperbola and the lines joining its centre tothe points corresponding to t 1 and – t 1 . [IIT-1982]Sol. We have to find the area of the region bounded bythe curve x 2 – y 2 = 1 and the lines joining the centrex = 0, y = 0 to the point (t 1 ) and (– t 1 )yP(t 1 )–1CA1 NRequired areat⎡1 – te + e 1 ⎤⎢2 ⎥= 2 ⎢area of ∆PCN – ⎥⎢ ∫y dx⎥⎢1⎥⎣⎦⎡ ⎛t1– t⎞⎛⎞ ⎤= 2 ⎢ ⎜+1 t1– tt1 e e⎟⎜e – e1⎟ ⎥⎢∫ 1dx– y . dt2⎥⎣ ⎝ 2 ⎠⎝2 ⎠ dt1 ⎦⎡ 2t1–2tt⎛ ⎞ ⎤= ⎢⎜ ⎟ ⎥⎢ ∫ 121 t – te – e e – e–dt8⎥⎣0 ⎝ 2 ⎠ ⎦xXtraEdge for IIT-JEE 10 SEPTEMBER 2011