Title 8_Hints & solutions.docx ✅

PRE-MEDICAL 38 PHYSICAL CHEMISTRY Serial No. MODULE-1 Page No. 1. Some basic concepts of Chemistry 2. Atomic structure 3. Chemical Equilibrium 4. Ionic Equilibrium 5. Thermodynamics & Chemical Energetics 6. Redox reactions 7. Behaviour of Gases 8. Hints and Solutions
PRE-MEDICAL 38 SOME BASIC CONCEPTS OF CHEMISTRY EXERCISE-I 3. (1) 1 g of O = 1 16 ×N A atoms (2) 1 gm of O 2 = 1 32 × 2N A atoms (3) 1 gm of O 3 = 1 48 ×3 N A atoms, Hence all have the same number of atoms 4. Number of atoms in 0.5 gm atoms of nitrogen = 0.5 N A 8 gm of oxygen = 8 16 = 0.5 N A atoms of oxygen 6. 1 mol. of NH 3 contains 7×N A neutrons 1.7 gm of NH 3 = 1.7 17 = 0.1 mol  0.1 mol contains  0.1 × 7 × N A neutrons 7. Moles of O 2 = 5.6 22.4 = 1 4  1 mol of O 2 contains 2N A atoms  1 4 mol of O 2 contains A2N 4 = AN 2 atoms 8. Q 1 mol of O 3 contains 3N A atoms of oxygen 8 gm O 3  8 48 moles = 1 6 mol  1 6 mol of O 3 contains  1 6 ×3N A = AN 2 17. Let the atomic weight of A and B are a & b wt of 0.05 moles of B 2 A 3 = 9 and wt of 0.1 mole of B 2 A = 10 gm Then wt of 1 moles of B 2 A 3 = 180 and wt of 1 mole of B 2 A = 100  wt of 1 mole of compound = mol wt. Mol. wt of B 2 A 3 : 2b + 3a = 180 ……(i) Mol. wt of B 2 A : 2b + a = 100 …...(ii) Solve eq. (i) & (ii) and get value of a &b 23. Mol. wt. = 2 × V.D.
PRE-MEDICAL 38 = 2 × 70 = 140 Mol. wt. of (CO) x = 28 x Hence 28x = 140 24. mol. wt. = 2 × V.D.  2×11.2 = 22.4 Moles of gas = 2.4 22.4 [ 1 mole occupies = 22.4]  volume of gas = 2.4 L 25. Moles of gas  7 1.1210(mL) 22400(mL)   Number of molecules of gas = 7 1.1210 22400   × N A 26. Moles of sugar (C 12 H 22 O 11 ) = 1.71 342  In one mole C 12 H 22 O 11 number of C atoms = 12 N A  1.71 342 mole C 12 H 22 O 11 number of C atoms = 1.71 342 ×12 N A 27. Moles of Ba(NO 3 ) 2 = M×V (Lit) = 0.1 × 1 –3 = 10 –4 mol  1 mol Ba(NO 3 ) 2 contains 3 mol ions  1 mol Ba(NO 3 ) 2 contains 3 ×10 –4 mol ions 43. % of S = 1at.wt.ofS100 minimummolecularmass  3.4 = 132100 m  Molecular mass (m) = 32100 3.4  44. % of metal = 1at.wt.ofmetal Minimummolecularweight  ×100 0.25 = 159100 m  M = 59100100 0.25  = 33600 45. wt. of nitrogen in caffine = 19428.9 100 
PRE-MEDICAL 38 Number of atom in one molecule = 19428.9 10028   ×2 47. C 8 H 18 + 25 2 O 2  8CO 2 + 9H 2 O d = 0.8 gm/mL Volume of petrol (C 8 H 18 ) = 1.425 L = 1425 mL wt. of C 8 C 18 = 0.8×1425 = 1140 gm mol. wt. of C 8 H 18 = 114, Moles of C 8 H 18 = 1140 114 = 10  1 mol C 8 H 18 requires 25 2 mol of O 2  10 mol C 8 H 18 will require = 25 2 ×10 mol of O 2 48. Moles of Al = 9 27 = 1 3 Moles of O 2 = 311 223 Weight of O 2 = 311 223 ×32 50. CO 2 (g) + C(s)  2CO(g) Initially 1 L After reaction (1 – x)L 2×L 1 – x + 2x = 1.4 x = 0.4 L Hence volume of CO 2 = 1 – 0.4 = 0.6 L & volume of CO = 2 × 0.4 = 0.8 L 51. CO 2 + C  2CO (Red hot) 1 vol. 2 vol. 26 CC 52 CC 53. C 2 H 2 + 5 2 O 2  2CO 2 + H 2 O   1 vol. 5 2 vol.  1 L C 2 H 4 requires  5 2 L of O 2