May 2011 - Career Point

60 cmK 1 K 2VB5. A solid sphere of copper of radius R and a hollowsphere of the same material of inner radius r and outerradius A are heated to the same temperature andallowed to cool in the same environment. Which ofthem starts cooling faster ?[IIT-1982]Sol. Since the temperature and surface area is same,therefore the Energy emitted per second by bothspheres is same.We know that Q = mc∆TSince Q is same and c is same (both copper)1∴ m ∝∆TMass of hollow sphere is less∴ Temperature change will be more.∴ Hollow sphere will cool faster.mCHEMISTRY6. 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 0.87778 1= × L = 244.58 L0.877 1000And molar volume of benzene in vapour phse78 2750= × L = 244.58 L0.877 1000Density 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 phase92 7720= × L = 819.19 L0.867 1000Using the ideal gas equation,PV = nRTAAt T = 20ºC = 293 K0 nRTFor benzene, P = P B =V1 × 0.082×293== 0.098 atm244.58= 74.48 torr (Q 1 atm = 760 torr)Similarly, for toluene,0 nRTP = P T =V1 × 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,' PB74 .48×0.457x B = == 0.74P 46.0M7. 0.15 mol of CO taken in a 2.5 L flask is maintained at705 K along with a catalyst so that the followingreaction takes placeCO(g) + 2H 2 (g) CH 3 OH(g)Hydrogen is introduced until the total pressure of thesystem is 8.5 atm at equilibrium and 0.08 mol ofmethanol is formed. Calculate (a) K p and K c and (b)the final pressure if the same amount of CO and H 2 asbefore are used, but with no catalyst so that thereaction does not take place. [IIT-1993]Sol. We haveCO(g) + 2H 2 (g) CH 3 OH(g)t = 0 0.15 molt eq 0.15 mol – x ( n H 2) 0 – 2x xIt is given that 0.08 mol of CH 3 OH is formed atequilibrium. Hencen CH 3 OH = x = 0.08 moland n CO = 0.15 mol – x = 0.07 molFrom the total pressure of 8.5 atm equilibrium, wecalculate the total amount of gases, i.e. CO, H 2 andCH 3 OH at equilibrium.pV (0.08mol / 2.5L)n total = = RT– 1 −1(0.082atm L K mol )(705K)= 0.3676 molNow, the amount of H 2 at equilibrium is given asn H 2= n total – n CO – n CH 3 OH= (0.367 – 0.07 – 0.08) mol = 0.2176 molXtraEdge for IIT-JEE 9 MAY 2011