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