TOPIC 3 ANSWERS & MARK SCHEMES QUESTIONSHEET 1 MOLAR SOLUTIONS (1)

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1 QUESTIONSHEET 1 MOLAR SOLUTIONS a) Molar concentration The number of moles of solute dissolved in 1 dm 3 of solution Molar solution One which contains 1 mol of solute in 1 dm 3 of solution b) (i) (HCl) = 36.5 Concentration = 146/36.5 = 4.00 mol dm -3 (ii) (KI) = 166 Concentration = 300/166 = 1.81 mol dm -3 c) (i) Concentration = = 5.92 g dm -3 (ii) (H 2.2H 2 O) = 126 Concentration = 5.92/126 = mol dm -3 d) 1 M Na 2 contains 106 g dm M Na 2 contains = 5.30 g dm -3 in 250 cm 3 there is 5.30/4 = g Na 2 e) Number of moles in solution = molarity volume (in dm 3 ) = = (NaOH) = 40 Mass = = 1.96 g

2 QUESTIONSHEET 2 PREPARATION OF A STANDARD SOLUTION a) (i) A substance which can be weighed out directly to give a solution of known concentration (ii) Must be readily available in a very pure state Must be stable / must not deteriorate on standing Must be non-hygroscopic / must not absorb water vapour Should have a high so that a substantial amount is weighed out / so that weighing errors are negligible (Accept any 3) b) (i) For accuracy (ii) To prevent the sodium carbonate from soiling / damaging the balance pan if it was spilt (iii) To allow for particles of sodium carbonate which adhered to the weighing bottle / which were not transferred (iv) Tap water contains dissolved substances which could affect the titration (v) To avoid errors due to some sodium carbonate solution remaining in the beaker (vi) To avoid overfilling the flask (vii) When the bottom of the liquid meniscus coincided with the graduation mark (viii)to get a uniform / homogeneous solution Or to avoid a concentration gradient in the flask (ix) To give the air bubble a chance to move up and down / to ensure that the solution became properly mixed (x) Discarded the solution / started again

3 QUESTIONSHEET 3 a) H 2 (aq) + 2KOH(aq) K 2 (aq) + 2H 2 O SIMPLE VOLUMETRIC CALCULATIONS n (H 2 ) = n (KOH) / 2 = 20.0(10-3 )0.5 / 2 = mol H 2 Volume required = 5(10-3 ) / 0.2 = 25 x 10-3 dm 3 = 25.0 cm 3 b) NaOH(aq) + HCl(aq) NaCl(aq) + H 2 O n (NaOH) = n (HCl) = 19.8(10-3 ) 0.15 = mol NaOH c (NaOH) = 2.97(10-3 ) / 25.0(10-3 ) = mol dm -3 = mol dm -3 c) CH 3 COOH(aq) + NaOH(aq) CH 3 COO - Na + (aq) + H 2 n (NaOH) = n (CH 3 COOH) = 0.148/60 = mol Volume required = 2.47(10-3 )/ = dm 3 = 24.9 cm 3 d) Na 2 (aq) + 2HCl(aq) 2NaCl(aq) + CO 2 (g) + H 2 O n (Na 2 ) = n (HCl)/2 = 19.6(10-3 ) 0.103/2 = mol m (Na 2 ) = 1.01(10-3 ) 106 = g

4 QUESTIONSHEET 4 CALCULATIONS REQUIRING MOLAR CONCENTRATION OF PRIMARY STANDARD a) n (Na 2 ) = 1.547/106 = mol c (Na 2 ) = (10 3 )/250 = mol dm -3 Na 2 (aq) + H 2 (aq) Na 2 (aq) + CO 2 (g) + H 2 n (H 2 ) = n (Na 2 ) = 25.0(10-3 ) = mol c (H 2 ) = 1.46(10-3 )/24.6(10-3 ) = mol dm -3 b) (i) Solid sodium hydroxide is deliquescent / absorbs water vapour from the air Hence solutions will be more dilute than calculated (ii) n (H 2.2H 2 O) = 14.6/126 = mol c (H 2.2H 2 O) = 0.116(10 3 )/250 = mol dm -3 H 2 (aq) + 2NaOH(aq) Na 2 (aq) + 2H 2 n (NaOH) = n (H 2.2H 2 O) 2 = 25.0(10-3 ) (0.464) 2 = mol c (NaOH) = /24.1(10-3 ) = mol dm -3 (iii) H 2 (aq) + 2NaOH(aq) Na 2 (aq) + 2H 2 n (H 2 ) = n (NaOH)/2 = 20.7(10-3 ) 0.963/2 = mol c (H 2 ) = 9.97(10-3 )/2.00(10-3 ) = 4.98 mol dm = 488 g dm -3

5 QUESTIONSHEET 5 DILUTION QUESTIONS a) A very large volume of titrant would otherwise be required / the burette would have to be refilled during the titration b) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 n (HCl) = n (NaOH) = 23.6(10-3 ) = mol c (diluted HCl) = 2.34(10-3 )/25.0(10-3 ) = mol dm -3 c (concentrated HCl) = (10 3 )/10.0 = 9.36 mol dm -3 c) (i) CH 3 COOH(aq) + NaOH(aq) CH 3 COO - Na + (aq) + H 2 O (l) n (CH 3 COOH) = n (NaOH) = 25.5(10-3 ) = mol c (diluted CH 3 COOH) = 2.60(10-3 )/25.0(10-3 ) = mol dm -3 c (original CH 3 COOH) = 0.104(200)/25.0 = mol dm (350)/10 3 = mol per 350 cm (60) = 17.5 g per 350 cm 3 (ii) From (i), c = mol dm (1.14) = mol per 1.14 dm (60) = 56.9 g per 1.14 dm 3 (iii) n (NaOH) = n (CH 3 COOH) = 0.832(25.0)/10 3 = mol Volume required = /0.102 = dm 3 = 204 cm 3

6 QUESTIONSHEET 6 WATER OF CRYSTALLISATION a) (i) H 2 (aq) + 2NaOH(aq) Na 2 (aq) + 2H 2 n (H 2 ) = n (NaOH)/2 = 21.1(10-3 ) (0.100)/2 = mol in 25 cm 3 n (anhydrous H 2 ) in 1 dm 3 = 1.055(10-3 ) (10 3 )/25 = mol (ii) (H 2 ) = 90 m (anhydrous H 2 ) in 1 dm 3 = (90) = 3.80 g (iii) m (hydrated H 2 ) in 1 dm 3 = 1.33(4) = 5.32 g m (H 2 O) in 1 dm 3 = = 1.52 g (iv) n (H 2 O) in 1 dm 3 = 1.52/18 = mol (v) n (H 2 ) : n (H 2 O) = : = 1 : 2 Or no. of moles of water of crystallisation = 2 b) H 2 (aq) + K 2 (aq) K 2 (aq) + H 2 O + CO 2 (g) n (K 2 ) = n (H 2 ) = 30.0(10-3 ) (0.0125) = mol in 25 cm 3 in l dm 3 there are 3.75(10-4 )(40) = mol K 2 (K 2 ) = 138 in 1 dm 3 there are 0.015(138) = 2.07 g K 2 Hence, mass of water of crystallisation = = g 0.405/18 = mol H 2 O with mol K 2 x = /0.015 = 1.5

7 QUESTIONSHEET 7 a) (i) Absorption of carbon dioxide from the air (ii) Ca(OH) 2 (s) + CO 2 (g) Ca (s) + H 2 PERCENTAGE PURITY b) (i) Ca would have reacted with HCl(aq) Hence too much HCl(aq) would have been used in the titration (ii) Ca(OH) 2 (aq) + 2HCl(aq) CaCl 2 (aq) + 2H 2 n (Ca(OH) 2 ) = n (HCl)/2 = 20.8(10-3 ) 0.210/2 = mol (Ca(OH) 2 ) = 74 m (Ca(OH) 2 ) = 2.18(10-3 ) 74 = g Purity = 0.162(100)/0.204 = 79.4% c) Na 2 (aq) + 2HCl(aq) 2NaCl(aq) + CO 2 (g) + H 2 Note NaCl does not react with hydrochloric acid. n (Na 2 ) = n (HCl)/2 = 18.6(10-3 )0.105/2 = mol in 25 cm mol Na 2 in 250 cm 3 (Na 2 ) = 106 m (Na 2 ) = 9.77(10-3 )106 = 1.04 g m (NaCl) = = 4.03 g Purity = 4.03(100)/5.07 = 79.5%

8 QUESTIONSHEET 8 TEST QUESTION I a) Let the acid be represented by HA; then, since it is monobasic, n (HA) = n (NaOH) = 25.0(10-3 )0.1 = mol in 25.0 cm 3 there were 10( ) = mol HA in 250 cm 3 (HA) = 1.15 / ( ) = g mol -1 b) Mg (s) + 2HNO 3 (aq) Mg(NO 3 ) 2 (aq) + CO 2 (g) + H 2 (Mg ) = 84 n (Mg ) = 1.00/84 = mol n (HNO 3 ) = 2n (Mg ) = 2(0.0119) = mol Volume required = /0.40 = dm 3 = 59.5 cm 3 c) (i) Ca (s) + 2HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 n (Ca ) = n (HCl)/2 = 10(10 3 )/2 = 5000 mol (Ca ) = 100 m (Ca ) = 5000 (100) = 500,000 g = 500 kg (ii) Reason 1 Ca is naturally occurring and hence cheap Reason 2 Any excess NaOH would pollute the environment but Ca being insoluble would not do so

9 QUESTIONSHEET 9 TEST QUESTION II a) (i) HCl(aq) + NaOH(aq) NaCl(aq) + H 2 (ii) n (NaOH) = 0.5(19.2/1000) = 9.6 x 10-3 mol n (HCl) = n (NaOH) = 9.6 x 10-3 mol c (HCl) = 9.6 x 10-3 / (25.0/1000) = M b) (i) Na 2 (aq) + 2HCl(aq) 2NaCl(aq) + H 2 O(l) + CO 2 (g) (2) Award for correct reactants and products and for correct balancing. (ii) n (HCl) = 0.384(19.1/1000) = x 10-3 mol n (Na 2 ) = n (HCl) / 2 = x 10-3 / 2 = x 10-3 mol c (Na 2 ) = x 10-3 / (25/1000) = M (iii) Moles Na 2 in 25 cm 3 = x 10-3 moles Na 2 in 1 dm 3 = (3.667 x 10-3 / 25) x 1000 = Mass Na 2 in 1 dm 3 = x 106 = 15.6 g percentage Na 2 = (15.6/20.0) x 100 = 78% c) 500 g Econosoda contains 500(78/100) = 390 g Na g Deluxwash contains 500(98/100) = 490 g Na 2 Econosoda costs 175/390 = 0.45p per gram Na 2 (½) Deluxwash costs 245/490 = 0.50p per gram Na 2 (½) Econosoda is the better value for money

10 QUESTIONSHEET 10 TEST QUESTION III a) HCl(aq) + NaH (aq) NaCl(aq) + H 2 O(l) + CO 2 (g) b) (i) m (NaH ) = 5.00(85/100) = 4.25 g n (NaH ) = 4.25/84 = mol c = / (250/1000) = mol dm -3 (ii) n (NaH ) = (25/1000) = 5.06 x 10-3 mol n (HCl) = n (NaH ) = 5.06 x 10-3 mol V (HCl) = 5.06 x 10-3 / 0.15 = dm 3 = 33.7 cm 3 c) n (H 2 ) = 2(10/1000) = 0.02 mol 2 mol NaH 1 mol H 2 n (NaH ) = 0.02 x 2 = 0.04 mol m (NaH ) = 0.04 x 84 = 3.36 g one 5.0 g tablet is sufficient d) 1 In a suspension particles have a larger surface area than in a tablet reaction with acid occurs more rapidly 2 1 mol of Mg(OH) 2 produces 2 mol of OH - (aq) Mg(OH) 2 is twice as effective as NaH 3 Mg(OH) 2, unlike NaH, does not produce CO 2 which can cause flatulence Maximum 4 marks

11 QUESTIONSHEET 11 EXPERIMENT TO FIND THE EQUATION FOR A REACTION a) (i) moles of iron used = 0.512/55.8 = (ii) moles of copper formed = 0.614/63.5 = (iii) ratio of moles of copper formed: moles of iron reacted = / = 1.05 (iv) Hence write the equation for the reaction Fe + Cu Fe + Cu Or Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) b) (i) Any three of: only carried out once not all iron reacted solid not dry solid lost when liquid removed (ii) (iii) Any one of: value is close to 1:1 only one value Any three of: repeat and average use finer iron filings filter off solid dry to constant mass Quality of language: at least two sentences in which the meaning is clear

12 QUESTIONSHEET 12 ESTIMATING THE CONCENTRATION OF SALTS IN SEA WATER a) Sample x 1000/25 = 28.1 g Sample x 1000/25 = 26.0 g Three significant figures b) Wide variation, result unreliable c) Incomplete evaporation of water Loss of residue due to spitting d) Heat to constant mass Heat in a flask Quality of language: at least two sentences in which the meaning is clear

13 QUESTIONSHEET 13 EXPERIMENT TO FIND THE EQUATION FOR A REACTION DECOMPOSITION OF POTASSIUM HYDROGEN CARBONATE a) Loss in mass = = g % Loss in mass = x100 = 30% b) i. % Loss in mass = 62 x100 = 31% 200 ii. % Loss in mass = 18 x100 = 9% 200 iii. % Loss in mass = 46 x100 = 23% 200 iv. % Loss in mass = 106 x100 = 53% 200 v. % Loss in mass = 50 x100 = 25% 200 c) A. 2KH (s) K 2 (s) + H 2 O(l) + CO 2 (g) d) Any three of: Only carried out once. Small mass % error high. Further decomposition. Spitting. (3) e) Reliable because of the close agreement between experimental and theoretical values, or unreliable because there is only one value. f) Any three of: Repeat. Use larger mass. Use an oven for temperature control. Put on lid (3) Quality of language: at least two sentences in which the meaning is clear

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