0.350 (= mol) 1 32

Transcription

1 1. PRACTICE EXAMINATION QUESTIONS MARK SCHEME 1. (a) (i) Avogadro s number/constant of molecules/particles/species / [Not atoms ] Or same number of particles as (there are atoms) [Not molecules] in 1.(00)g of 1 C 1 (ii) Moles O = (= mol) 1 3 = 9 ( ) 1 [Accept answers via 4 separate mole calculations] = mol [answer to 3+ sf] 1 [Mark conseq on errors in M1/M] (iii) Moles of nitroglycerine = (= mol) 1 [Mark conseq on their moles of O ] M r of nitroglycerine = 7 or number string 1 Moles of nitroglycerine = = (g) [answer to 3+ sf] [If string OK but final answer wrong then allow M6 but AE for M7] [Mark conseq on error in M r ] [Penalise wrong units] [Penalise sig. fig. errors once only in whole question] (b) pv = nrt or pv = p = nrt V = mrt or p = V nrt V = or 7980 or [ignore s.f.] 1 1 units = Pa or kpa or MPa (as appropriate) 1 [If error in conversion from Pa, treat as a contradiction of the units mark] [If transfer error, mark conseq but penalise M] [If data from outside of above used, penalise M and M3] [If pv expression incorrectly rearranged, penalise M and M3] [if T = 1373 K used, penalise M] [11] Mill Hill High School 1

2 . (a) Na Cl O (39) 0.9(38).8() Hence: Accept backwards calculation, i.e. from formula to % composition, and also accept route via M r to 3; 35.5; 48, and then to 1:1:3 [If % values incorrectly copied, allow M1 only] [If any wrong A r values/atomic numbers used = CE = 0} (b) 3Cl + 6NaOH 5NaCl + NaClO 3 + 3H O 1 [3] 3. (penalty for sig fig error =1mark per question) (a) (i) moles KNO 3 = 1.00/101.1 = (mol) 1 (ii) pv = nrt or n = pv/rt 1 pv moles O = n = = RT = (mol) 1 (mark answer first check back if wrong) (transcription error lose M3, mark M4 conseq on error) (if untraceable figures used M3=M4=0) (if wrong temp conversion lose M3 conseq M4) (if n = RT/pV CE, lose M3 and M4) (b) (i) simplest/lowest ratio of atoms of each / element/s in a compound / substance / species / entity / molecule 1 (ii) K N O KNO 3 (M3 tied to M), (M3 can be transferred from equation if ratio correct but EF not given) (if calc inverted, lose M and M3), (if used At N 1 / wrong No for Ar then CE, lose M and M3) (if % of O missing, award M only) (c) KNO 3 KNO + O or fractions/multiples 1 (accept KNO 3 K N O 4 + O ) (do NOT accept Y in equation) [10] Mill Hill High School

3 4. (penalty for sig fig error =1mark per question) Mr (Mg(NO 3 ) = 58(.3) (if At N used, lose M1 and M) 1 moles Mg(OH) = (conseq on wrong M ) (answer to 3+ s.f.) 1 moles HCl = = or (mol) (process mark) vol HCl = = (cm 3 ) (unless wrong unit) 1 1 (if candidate used or lose M) (just answer with no working, if in range = (4). if, say, 34 then =()) (if not :1 ratio, lose M3 and M4) (if work on HCl, CE = 0/4) [4] 5. A1+ 3CuCl A1C Cu; (accept multiples/fractions) OR A1+ 3Cu + A Cu; 1 [1] 6. (a) moles HN0 3 = 175 X = (0.65 mol); 1 moles Pb(NO 3 ) = ½ 0.65 = (0.131 mol); 1 M r Pb(NO 3 ) = 331(.); 1 mass Pb(NO 3 ) = 331. x 0.131=43.5 g; 1 (accept ) (M1 & M are process marks. If error in M1, or in M, do not mark M4 consequentially, i.e. do not award M4) (if atomic numbers used in M3, do not award M4) (b) (i) pv = nrt; 1 4 pv n = = ; RT = 3.61 X 10 3 ; 1 (If pressure not converted to Pa, max ) RT (If n = used = CE; M = M3 = 0) pv (ii) molesn0 = 4/ ; 1 [mark is for use of 4/5] = OR ; 1 M r NO = 46; 1 massno = l0 3 = (g) OR (g); 1 ( if atomic numbers used, M3 = M4 = 0) [11] Mill Hill High School 3

4 7. (a) Na + NH 3 NaNH + H (or multiples) 1 (b) (i) Simplest ratio of atoms of each element in a compound / substance / species / entity / molecule 1 (ii) Mg N O (4) (0.675) MgN O 6 (Mark M1 first. If any wrong A r used = CE = 0) (Accept Mg(NO 3 ) for M3 if above working shown) [4] 8. (a) (i) 100 l = (mol) accept 5 10 / (ii) = / l0 (mol) only 1 (iii) (mol) 1 Mark conseq on (ii) (iv) = (mol) 1 Mark conseq on (i) & (iii) (v) ½ = (mol) If used Mark conseq on (iv) l3(.l) or (.l) 1 Mark for M r =. g or 1.83 g 1 (b) pv = nrt 1 n = = (mol) 1 17 Note pv T = = 3 nr = (K) 1 Mark conseq on moles Sig. fig. penalty - apply once if single sf given, unless calc works exactly [11] Mill Hill High School 4

5 9. Ideal gas equation: pv = nrt Calculation: n = pv/rt = ( ) mark for volume conversion fully correct 6 = (mol) range to M r = m/n =.304/ = range min s.f. conseq If V wrong lose M; p wrong lose M3; inverted lose M3 and M4 10. (a) M1 % by mass of H = 7.7(0)% [5] M mol H = 7.70 / 1 = 7.70 mol C = 9.3 / 1 = 7.69 M3 (ratio 1:1 ) CH Credit variations for M 78 = 6 and = Correct answer = 3 marks 78 (b) (CH has empirical mass of 13 and = 6 ) C6 H 6 13 Correct answer 1 mark 4 [4] 11. 3/ CE = 0 if inverted or multiplied tied to M1 3.8() 10 3 [-5 sig figs] 1. (a) (simplest) ratio of atoms of each element in compound (b) % oxygen = 39.5% Na 8.4/3 Cr 3.1/5 O 39.5/16 = 1.3 = =.47 (:1:4) so empirical formula = Na CrO 4 4 If % oxygen not calculated, only M available; if A r values wrong, only M1 available [] [4] Mill Hill High School 5

6 13. (a) 0 40 Mr (TiC1 4 ) = 190 moles TiC1 4 = 0 4 = mass = = (b) moles NaOH = moles HCl = 0.1 vol NaOH = moles conc = 0 1 = 109cm moleshc1 (c) moles H = = 0 06 (allow conseq) V = nrt or PV = nrt p = = m 3 etc (if chemical error on moles H, max) () 4 [11] (a) moles of S = ( = 0.315) allow mass of sodium sulphide i.e. 4.4 g or 4.37 g (accept 4.3 g or 4.33 g) (no sig.fig. penalty) (ignore units) (b) moles of hydrogen sulphide (i.e or mol) moles of hydrogen gas needed = moles of H S volume: i.e. 359 or 3590 cm 3 (360 or 3600 if used) 3 answer must be in cm 3 ie 4500, not 4.5 / no s.f. penalty / ignore units allow PV = nrt method with remembered value of R [5] Mill Hill High School 6

7 15. (a) Ideal gas equation law 1 4 PV (b) Moles of X: n = = RT = to , min sig figs RT If write n = zero here, but can score M r PV m Relative molecular mass of X: M r = n = to 6.5 (c) % oxygen = 51.6 () C =38.7 / 1 H = 9.68 / 1 O = 57.6() / 16 = 3.3 = 9.68 = : 3: 1 CH 3 O If no % O or if wrong A r used then max 1 Correct empirical formula earns all three marks 3 6 (d) ( CH3 O) = C H 6 O 1 31 [9] mass (a) Moles HCl = = (= 0.537) M r Concentration = 0.5 =.15 (mol dm 3 ) mass Conseq on correct M r min d.p..14 to.15 Ignore units A.E. lose one mark 3 Mill Hill High School 7

8 (b) (i) = (mol) to (ii) Conseq 3 = (mol) (iii) Conseq = 138 (iv) Relative atomic mass of M: = = 39 Identify of M: Potassium or K or K + Conseq If 78 = M r then M = selenium 6 [9] 17. (a) mass of 1mol L = 4 or mass of 1atom must show working = Ignore wrong units NB answer only scores 1 (b) equal Or same or 1:1 1 (c) PV PV = nrt (or n = ) RT = = 1.39 Allow to ignore units even if incorrect answer = 1.4 loses last mark 3 Mill Hill High School 8

9 1000 (d) 0.73 =.93 mol.dm 3 50 OR M, mol/dm 3, mol.l 1 allow.98 to.93 Note unit mark tied to current answer but allow unit mark if answer =.9 or 3 5 (e) (i) moles H SO 4 = 1.4 = If use m 1 v 1 = m v scores 3 if answer is correct otherwise zero moles NH 3 in 30.8 cm 3 = = Mark is for CE if not used moles of NH 3 in 1 dm = 0.60 =.01 (mol dm 3 ) 30.8 Allow.010 to.015 No units OK, wrong units lose last mark (ii) moles (NH 4 )SO 4 = moles H SO 4 = Allow consequential wrong moles in part (i) if clear H SO 4 =(NH 4 )SO 4 Wrong formula for (NH 4 )SO 4 CE=0 M r (NH 4 )SO 4 = 13.1 Allow (13) mass = moles M r = = 4.10 if moles of (NH 4 )SO 4 not clear CE (g) wrong unit loses mark Allow (f) Mg 3 N + 6H O 3Mg(OH) + NH 3 Formulae Balanced equation [16] Mill Hill High School 9

10 18. (a) average mass of an entity 1 1 mass of 1atom of C (b) simplest ratio of atoms of each element in a compound 1 (c) (i) %O = C:H:O = : : = 1.96: 1 :.93 = C HO 3 (ii) C 4 H O 6 4 (d) (i) mass 1000 = = 6.5 M r scores one (ii) pv= nrt nrt V = = P = 3.10 m 3 (allow consequential marking but if factor of in missing max = ) 6 [13] 19. Avogadro s number of molecules allow 6 to molecules OR No. of atoms in 1g of 1 C 1 0. (a) Avogadro s 1 (b) (i) pv=nrt [1] allow {V= nrt (ii) V= P nrt m etc., pv= RT P Mr = = m 3 (allow to 0.05) Mill Hill High School 10

11 (iii) allow use of 100 for P if ans given as 4.8 dm 3 pv n= RT = (treat 500 as AE 1 other values l s mar 0.005/1000 loses 1 st two marks = 1.10mol (allow 1.1 to 1.11) Mr (CO) = 44 treat wrong Mr as AE 1 mass = = 48.5 (g) (allow 48 49) PV (note can use calculation involving 1 1 and.4 dm 3 instead of 1 st mark - T requires same accuracy for full marks) 100 (c) moles HC1 = 5.0 = 0.5(0)(mol) 1000 moleshc1 moles H = = 0.5 (mol) (if no factor of CE = O from here) mass H = 0.5 = 0.5(0) (g) 3 (d) (i) %O = 44.4 (if incorrect %O, AE 1) (if %O omitted can score max 1 for FeC ) ratio Fe:C:O = = 1 : : : : (or 50) Fe C O 4 (ii) CO (mark independent of d (ii)) (if use At, CE) 7 4 [15] (a) moles of NH 3 = 17.0 moles of HNO 3 = moles of NH 3 = volume = = dm 3 4 Mill Hill High School 11

12 (b) (i) PV = nrt (ii) n PV = = RT = 1.9 mol Moles of ammonium nitrate mol 1.5 Mass of ammonium nitrate mass = = 10 g 6 [10] [[1] = (a) mass ratio of O :KClO 3 = 96 / 45. ( = 0.39) or moles of O = 3/ or equivalent mass of O = = 0.576g (b) moles O = 1.00/4.00 ( = mol) moles KClO 3 = /3 moles O = /3 = (0.078 mol) mass KClO 3 = = 3.4lg 3 give one mark for pv = nrt to get n for O, then second and third mark as above or /3 mole of KClO 3 1 Mole of O 81.7g 4 dm 3 O g = 3.4lg 1 dm 3 O penalise other than 5 sig. figs. once in (d)(i) and (ii) penalise missing or wrong units once in (d)(i) and (ii) [] [5] 4. (a) Mass of each element in the compound 1 (b) Number of atoms of each element in a molecule 1 (c) (i) Mr Ba(NO 3 ) = 61 5 moles Ba(NO 3 ) = = moles gas = = nrt V = = p = m 3 (ii) moles HCl = = vol HCl = = 0.03 dm 3 (or 3 cm 3 ) 7 1. [9] Mill Hill High School 1

13 5. percentage by mass of oxygen = 38.0% ratio of elements = 4.9/1 :.4/1 : 16.7/14 : 38.0/16 = 3.6 :.4 : 1. :.4 empirical formula is C 3 H NO molecular formula is C 6 H 4 N O 4 4 [4] 6. (a) %H = B : H = : = : 5 : B H 5 (b) B 4 H 10 4 [4] 7. (a) mean mass of an entity (or molecule) 1 mass of 1 atom of 1 C 1 (b) = (allow 6.00 to 6.03) 1 [3] 8. (b) (i) %F = I:F = 17 = IF 5 = (ii) I + 5F IF 5 (or ½) 4 [4] Mill Hill High School 13

14 9. C.4/1 = 1.85 H 3.71/1 = 3.71 Br 74.05/79.9 = 0.97 ratio C:H:Br = :4:1 C H 4 Br empirical mass = mol formula = 15.8/107.9 C H 4 Br = C 4 H 8 Br must use % to justify answer or C (.4/100) 15.8 = i.e. 48/1 = 4 carbon atoms H (3.71/100) 15.8 = 8.01 i.e. 8/1 = 8 hydrogen atoms Br (74.05/100) 15.8 = i.e /79.9 = bromine atoms or C (48/15.8) 100 =.4% H (8/15.8) 100 = 3.71 % Br (159.8/15.8) 100 = 74.05% 3 [3] Mill Hill High School 14