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Syllabus – cum – Comp./XI/2013 – 14 1 S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084 Some basic concepts of chemistry (XI) Review No. 2 ( Solutions) Objective Questions Q1. What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53 g of metal?(a) V 2 O 3 (b) VO (c) V 2 O 5 (d) V 2 O 7 Sol. (c) Amount of O = 2.74-1.53 = 1.21Wt Moles Simplest Whole no. ratioM = 1.53 03.05153.1 103.03. 1 2 = 2O = 1.21 075.1621.1 5.203.075. 2.5 2 = 5= V 2 O 5 Q2. A compound having the empirical formula (C 3 H 4 O) n has a molar mass of 170 5. Themolecular formula of this compound is(a) C 3 H 4 O (b) C 6 H 8 O 2 (c) C 6 H 12 O 3 (d) C 9 H 12 O 3 Sol. (d) n = mass.EmpmassMolar Molar mass = [(12 3) + (4 1) + (16 1)]n = 170 5= (36 + 4 + 16) n = 170 556 n = 170n = 3n565170 M.F. = (E.F.) n = (C 3 H 4 O) 3 = C 9 H 12 O 3 Q3. What volume of hydrogen at NTP will be liberated when 3.25g of zinc completely dissolve indilute HCl? (At. mass of Zn = 65) (a) 1.12 litre (b) 11.20 litre (c) 2.24 litre (c) 22.40 litreSol. (a) Zn + 2HCl ZnCl 2 + H 2 1 mole 1 mole65 g 22.4 L3.25 g 25.3654.22 = 1.12 L Q4. Sucrose reacts with oxygen to yield CO 2 and H 2 O according to the reaction.C 12 H 22 O 11 +12O 2 12CO 2 + 11H 2 O.The number of molecules of CO 2 produced per gm of sucrose is (a) 2.11 10 22 (b) 6.02 10 22 (c) 1.76 10 21 (d) 9.29 10 23 Syllabus – cum – Comp./XI/2013 – 14 2 S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084 Sol. (a) C 12 H 22 O 11 12CO 2 1 mole produces 12 moles of CO 2 342 g produces 12 N A of CO 2 1 g produces 34210023.612 23 = 2.11 10 22 molecules Q5. 1.6 g of sulphur was burnt in the air to form SO 2 . The number of molecules of SO 2 introducedinto the air will be(a) 6.02 10 23 (b) 3.01 10 23 (c) 6.02 10 22 (d) 3.01 10 22 Sol. (d) S + O SO 2 1 mole of S produces 1 mole of SO 2 32 g of S produces N A molecules of SO 2 1.6 g of S produces 3210023.6 23 1.6 = 3.01 10 22 molecules Q6. Compute the value of xx = 9.4 g of phenol (C 6 H 5 OH) + 6.02 × 10 23 (wrong in review) molecules of phenol -0.2 mole of phenol (a) 0.9 mol (b) 9.2 g (c) 0.1 mol (d) 6.02 × 10 23 moleculesSol. (a) x = 9.4 g of phenol + 6.02 10 23 molecules - .2 molex = 944.9 moles + 1 mole - .2 mole= (.1 + 1 - .2) mole= .9 mole Q7. The number of moles of oxygen in one litre of air containing 21% oxygen by volume, instandard conditions, is(a) 0.186 mol (b) 0.21 mol (c) 0.0093 mol (d) 2.10 molSol. (c) 100 L of air has 21 L of O 2 1 L of air has 10021 = .21 L of O 2 Volume moles of O 2 (At S.T.P.)22.4 L 1.21 L 21.4.221 = .0093 moles Q8. A boy drinks 500 mL of 9% glucose solution. The number of glucose molecules he hasconsumed are [mol. wt. of glucose = 180](a) 0.5 × 10 23 (b) 1.0 × 10 23 (c) 1.5 × 10 23 (d) 2.0 × 10 23 Sol. (c) solutionof Volumesoluteof weightVW%9 100 mL glucose solution has 9g glucose Syllabus – cum – Comp./XI/2013 – 14 3 S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084 500 mL glucose solution has g5001009 = 45 g of glucose.Moles of glucose = 25.4118045 No. of Molecules = .25 6.023 10 23 = 1.5 10 23 Q9. List – I List – II(A) 0.5 mol (p) 92 g CH 3 CH 2 OH(B) 1.0 mol (q) 66g CO 2 (C) 1.5 mol (r) 54 g H 2 O(D) 2.0 mol (s) 15 g HCHO(E) 3.0 mol (t) 44g CH 3 CHOSol. Moles = massMolar massGiven (p) 6161212 92 = 2 (D)(q) 4466 = 1.5 (C)(r) 1854 = 3 (E)(s) 3015 = .5 (A)(t) 4444 = 1 (B) Q10. In the reaction, Al(s) + 6HCl (aq) 2Al 3+ (aq) + 6Cl - (aq) + 3H 2 (g)(a) 33.6 L H 2 (g) is produced regardless of temperature and pressure for every mole Al thatreacts (b) 67.2 L H 2 (g) at STP is produced for every mole Al that reacts (c) 11.2 L H 2 (g) at STP is produced for every mole HCl (aq) consumed. (d) 6 L HCl (aq) is consumed for every 3 L H 2 (g) produced.Sol. Using mole concept Q11. When 1 L of CO 2 is heated with graphite, the volume of the gases collected is 1.5 L. Calculatethe number of moles of CO produced at STP.(a) 2.111 (b) 4.2228 (c) 4.221 (d) 4.2214 Sol. (c) CO 2 (g) + C(s) 2CO(g)Initial volume 1 L OSuppose reacted xProduced 2x ∵ CO 2 2CO Syllabus – cum – Comp./XI/2013 – 14 4 S.C.O. 208 II floor, Sec-36 D Chandigarh, Ph.3012085, Mb.9872079084 At end 1 – x 2xTotal volume = 1 – x + 2x = 1.5 Given in Q’n 1 + x = 1.5x = .5 L Vol. of CO = 2 .5 = 1LMoles of CO = 4.221 Q12. Number of electrons present in 3.6 mg of NH 4+ are: (a) 1.2 × 10 21 (b) 1.2 × 10 20 (c) 1.2 × 10 22 (d) 2 × 10 -3 Sol. (a) 4 NH n = g414g106.3 3 = .2 10 -3 moles1 mole of NH 4+ has = 7 + 4 – 1 = 10 moles of e -1 s.2 10 -3 of NH 4+ has 3 1010210 moles of e -1 s= 2 6.023 10 23 10 -3 = 2 6.023 10 20 = 12.046 10 20 = 1.2 10 21 Q13. Caffeine has a molecular weight of 194. If it contains 28.9% by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is: (a) 4 (b) 6 (c) 2 (d) 3Sol. (a) 100 g of caffeine has 28.9 g of N194 g of caffeine has 1981009.28 = 57.22 g of N194 g = 1 molecule of caffeine = 57.22 g of N1 atom of N = 14 g 57.22 g = 1422.57 4 (No. of atoms of N) Q14. Four grams of hydrocarbon (C x H y ) on complete combustion gave 12 gram of CO 2 . What is theempirical formula of the hydrocarbon? [EAMCET (Medical) 2005](a) CH 3 (b) C 4 H 9 (c) CH (d) C 3 H 8 Sol. (d) Q15. The ratio of oxygen atoms to hydrogen atoms in (NH 4 ) 2 SO 4 is:(a) 1 : 1 (b) 2 : 1 (c) 1 : 2 (d) 4 : 1Sol. (c) 2184HO Q16. A certain compound has the molecular formula X 4 O 6 . If 10.0 g of the compound contains5.62g of X, the atomic mass of X is(a) 62.0 amu (b) 48.0 amu (c) 32.0 amu (d) 30.8 amu
