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And Meq. of M 2 CO 3 + Meq . of M 2´CO 3= Meq. of HClorx 2(1 − x)× 1000 +37 2m + 60× 1000= 0.5 × 44.4 ...(ii)Solving eq. (i) and (ii), we getm = 23and x = 0.41∴ Mass of M 2 CO 3 = x = 0.41 gand Mass of M 2´CO 3 = 1 – x = 0.59 gStep 5.Molecular massEquivalent mass of M 2 SO 4 =2110= = 552WM Meq. of M 2 SO 4 = 2 SO 455× 1000But, Meq. of M 2 SO 4 = Meq. of M 2 CO 3∴WM 2 SO 40.41× 1000 = × 10005537or WM 2 SO4= 0.6095 gStep 6.Molecular massEquivalent mass of M 2´SO 4 =2142= = 712WMMeq. of M 2´SO 4 =2 ´ SO 471× 1000But, Meq. of M 2´SO 4 = Meq. of M 2´CO 3∴WM2 ´ SO 42× 0.59× 1000 =71106× 1000or W = 0.7904 gM 2 ´ SO 4∴ Total mass of sulphates = WM 2 SO+ W4 M2´SO4= 0.6095 + 0.7904= 1.3999 g8. 0.9 g of a solid organic compound (molecular mass90), containing carbon, hydrogen and oxygen, washeated with oxygen corresponding to a volume of224 ml at STP. After combustion the total volume ofthe gases was 560 ml at STP. On treatment withpotassium hydroxide, the volume decreased to 112ml. Determine the molecular formula of thecompound.[IIT-1972]Sol. Given that,Mass of solid organic compound = 0.9 gMolecular mass of organic compound = 90∴ No. of moles of organic compound available0. 9= = 0.01 90Volume of O 2 taken = 224 mlVolume of O 2 used = 224 – 112 = 112 ml22400 ml O 2 = 1 mol.112∴ 112 ml O 2 = = 0.005 mol22400112∴ At STP, no. of moles of O 2 used = 22400= 0.005 molVolume of CO 2 obtained = 560 – 112 = 448 ml448∴ At STP, no. of moles of CO 2 used = 22400= 0.02 mol0.01 mol organic compound yields = 0.02 mol CO 2 .∴ 1 mol organic compound yields= 2 mol CO 2 or 2 mol C∴ The molecular formula of organic compound isC 2 H y O z .The reaction is :C 2 H y O z + 21O2 → 2CO 2 + 2yH2 OEquating no. of oxygen atoms,z + 1 = 4 + 2yorz = 3 + 2yMolecular mass of C 2 H y O z = 2 × 12 + y × 1 + z × 16⎛ y ⎞Hence, 2 × 12 + y × 1 + ⎜3 + ⎟ × 16 = 90⎝ 2 ⎠or y = 2and z = 3 + 2y = 4∴ The molecular formula of organic compound isC 2 H 2 O 4 .9. (a) Write the intermediate steps for each of thefollowing reactions.H O(i) C 6 H 5 CHOHC ≡ CH ⎯ ⎯ +3→ C 6 H 5 CH=CHCHO(ii)H +OHO CH 3(b) Show the steps to carry out the followingtransformations :(i) Ethylbenzene → benzene(ii) Ethylbenzene → 2-phenylpropionic acid[IIT-1998]Sol. (a) (i)C 6 H 5 CH(OH)C ≡ CH⊕C 6 H 5 CH = C = CHOH –H +C 6 H 5 CH – C ≡ CHOH 2+⊕C 6 H 5 CH – C ≡ CH–H 2 OTautomerismC 6 H 5 CH = C = CHOH C 6 H 5 CH = CHCHOXtraEdge for IIT-JEE 11 APRIL 2010

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