Chemistry Assignment-1 Atomic Structure (Solution)
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1 Chemistry Assignment-1 Atomic Structure (Solution) Phone No : /1101, 1
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9 XI (Strider Batch) Advance Assignment-1 CHEMISTRY Topic : Mole Concept 1. (C). (B). (A) HBO is a mono basic acid SOLUTIONS So M 1 = equivalent mass 4. (D) NaOH + FeCl Fe(OH) + NaCl m.e. of NaOH = me.e of Fe(OH) W mol.wt. 100 N = 1000 EFe(OH) E N = = = 0.4 N (C) Eq of Acid = eq of base n n = wt of A = 96 = (B) Let moles of FeO and FeO in the mixture is a and b respectively, then by POAC we get following two equation. a + b = 0.65 = 1....(i) a + b = 1...(ii) by solving (i) & (ii) we get a : b = 4 : 7. (B) Let normality of HC is N1 and HSO4 is N. M.e. of HC + M.e. of HSO4 = M.e of NaOH 5 N N = ( 1 ) N1 + N = (1) BY POAC Moles of C = moles of AgC 0 N N1 = 0.05 N N = 0.5 N 8. (A) M.e. of NH = 10 m.e. of NH = 10 m.mole of (NH4)SO4= 5 wt. of (NH4) SO4 = 5 5 wt. of (NH4) SO4 = 1 = 0.66 gm 1000 % of (NH4) S04 = = 94.8 % 9. (B) From given reactions Phone No : /1101, 67555/55 8
10 mmoles of hypo = mmoles of iodine = mmoles of Cu + ions = mmoles So mass of copper = gm So % of copper = 100% 51.0% (A, B, C, D) V 1 1+ V 0.5 Final molarity = (V + V ) (V1 + V) = V1 + V V V = V1 + V X V = 0.5 V1 V1 V 11. (A,B,C) MClx In each mole of MClx there are x moles of Cl - [Cl - ] = x 0.01 conc. of [M x+ ] = (A,B,D) (A) Molarity of second solution is = 10 d x 1M M (B) Volume = = 00 ml 001 (D) Mass of HS04 = 98 = 19.6 gm (A,B,C) For same number of molecules, number of moles should be same. 14. (A,B,D) When mixture is passed through hot graphite the following reaction will occur. CO (g) + C(s) CO (g) x ml x ml will formed Total volume of mixture = x + x = 160 x = 60 ml volume of CO = x = 40 ml 15. (D) One mole of SOI and O have same number of molecule. 16. (C) V α n at same temperature and Pressure. 17. False HMoO4 reduces itself therefore oxidise others and act as oxidising agent. 18. False For calculation of valency factor reactant product should be known. 19. (B) 0. (C) Moles of NaSO used = =. Moles of NaSO left unreacted = 6 -. = (A) No moles of NaSO would remain in step (i) so further reaction will stop because NaSO is required in excess.. to 4. On applying POAC for A & B respectively, we get 1 x na = x n AB (1) 5 x n B = 5 n AB () 5 Phone No : /1101, 67555/55 9
11 n n A B 5 By (1) & (), nab = = 4 5 Balanced equation: 4A + 5B AB5 q = 1.5 on applying POAC for X and Y respectively, we get 1 x nx = 1 x n XY () x n Y = x n XY (4) by () & (4), n Y = 1.5 x nx from () and (4) n n x Y nxy Balanced equation: X + Y XY 10 4 ny nxy 60 4 = mxy x100 1.g () C + O CO + CO W W x gm y gm POAC for C w x y w x y (1) 7 11 y x 4x 4y So now y x y x y x, 7: y POAC for O w x y 8 44 w x y. () (B). (B) 4. (A) 5. (C) Mass of oxygen in oxide = = 4.44 Eq. wt. of metal = mass of metal which combine with 8 gm of oxygen 5 = 8 = 9.00 gm 4.44 W W 6. (A) E E HPO4 Mg(OH) W W = 65. gm 7. (B) Equilise the m.e. of NaOH by m.e. of acid. 8. (C) M.e. of HPO = m.e. of NaOH = V 0.1 V = 100 ml HPO is monbasic acid Phone No : /1101, 67555/
12 9.(B) Number of moles of water produced = 0. moles 0. equivalent of acid should be reacted Wacid 0. Eacid = 49 E acid xa (D) Use formula (1 x )xm 1. (B) m NaCO = 1 n NaCO nno x NaCO A A and get answer.. (A) Let wg water is added to 16 g CHOH Molality = W W 500 W = xa (1 x )m A B W = 7 gm. (A) % (w/v) volume of NaOH = (B) Mole of solute = Mass of solvent = = 510 Molality = (A - p, s), (B - q, r, t) ; (C - p, q, s, t), (D r, t) (A) 4.1 gm = 4.1 mole HSO = 50 m mole 8 NaOH + HSO Na SO + HO m mole of NaOH requires = m mole of HSO = 100 = m mole of NaOH present (00 ml 0.5 N) Highest O.N. of S = + 6 (B) 4.9 gm = 4.9 mole = 50 m mole of HPO4 = 00 m mole O atom 98 Highest O.N. of P = + 5 (C) 4.5 gm = 4.5 = 50 m mole of HCO4 (di basic acid) 90 m mole of NaOH requires = 50 = 100 Highest O.N. of C = + 4 (D) 5. gm = 1 0 mole NaCO It do not react with NaOH and m mole of O atom = 1 0 = 150 Phone No : /1101, 67555/
13 6. (A - r); (B - s); (C - p); (D - q, r, (E - t) (A) 4.5, HSO4 means 4.5 moles of CaCO is present in 1000 g solvent. mass of solute = = 450 g mass of solution = 1000 ml Hence molarity = 4.5 M (B) Resultant molarity = M (C) mole fraction = (D) moles of NaOH in ltr = 4 = 8 mole moles of NaOH added = = 1 mole Molarit = 9 = 4.5 M mass of NaOH = 9 40 = 60 g 540 (E) % (w/w) NaOH = % % of carbon = m 6.5(10 m) On solving 8. wt of BaO = mole = = wt of CaO = 0.8 = mol. 56 BaO + HCl BaCl + HO CaO + HCl C acl + HO Number of moles of HCl required = MVL = V ml V = 9 ml. Phone No : /1101, 67555/55 1
14 NaOH 100ml,0.11MH SO4 9. X NaNO NaXO + NH (NH4)SO4 + HSO4 0.5M NaOH (excess) Now, Reactions involved for part (i) X XO - NJO - NH On balancing above equation in basic medium we get 4X + NO - + 7OH - 4XO - + NH + HO (1) NH + HSO4 (NH4)SO4 () (5mM) reacted Excess 48 ml of req HSO4 + NaOH NaSO4 + HO () ( Total m moles of HSO4 = 11) (excess) mmoles) =1mmoles m moles of X = =.6 At. wt. 40. Balance chemical equation are : Ca(PO4) + 8 Mg CaP + 8MgO CaP + 6HO Ca(OH) + PH PH + 4O PO5 + HO At. Wt. = 65 Ans. MgO + PO5 Mg(PO) moles of magnesium used = 0.8 moles moles of MgO formed = 0.8 moles moles of CaP formed 0.1 moles moles of PH formed = 0. moles moles of PO5 formed = 0.1 mole (limiting reagent) moles of Mg(PO) = 0.1 moles mass of Mg(PO) = 18. gram 41. If M is molar mass of (CH)x AlCly 0.64x mch 4 = 16=0.. M 0.64y and magcl = 14.5 = M dividing : x =, Also M = 15x y = 15x x y 4. Mass of NaSO solution = = 150 g 1164 g NaPO4 solution has 164 gm of NaPO4 100 g of HO will have 16.4 g of NaPO4 Vol of NaPO4 solution taken = ml Final volume of solution ml [ Na ] 1.5M x + 7. Phone No : /1101, 67555/55 1
15 4. Let mixture contain xg XH4 then (5.68 x)gxh6. x x moles of XH4 = and, moles of XH6 = M + 4 Also x 5.68 x = M+6 M+4 Mx Mx + = 5 M + 4 M + 4 Solving equation (i) and (ii) we geta M = 7.8. (M + 4)..(i)..(ii) 44. Volume of smallest cell = r = ( cm) = cm. mass of one smallest cell = g. Molar mass of mother cell = = amu. 45. Smallest volume of AgNO would be required when the entire mass is due to highest molecular weight constituent. Hence, for smallest volume, the wholemass should be of Bacl.HO. m mol of BaCl.HO = = 1.9 m mol. 44 m mol of AgNO required = 1.9 =.45. Volume of AgNO required =.458 = 16.8 ml (smallest) Largest volume of AgNO would be required when entire mass is due to lowest molecular weight constituent, i.e., NaCl. m mol NaCl = = m mol of AgNO required Volume of AgNO required = 5.18 = 4.18 ml (largest) Phone No : /1101, 67555/
15.0 g Fe O 2 mol Fe 55.8 g mol Fe = g
CHAPTER Practice Questions.1 1 Mg, O, H and Cl (on each side).. BaCl (aq) + Al (SO ) (aq) BaSO (s) + AlCl (aq).5 0.15 mol 106 g mol 1 = 1. g 15.0 g Fe O mol Fe 55.8 g mol Fe = 10.9 g 1 159.7 g mol FeO
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