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water displaced (mL) Time (seconds) 2 174 4 218 6 273 8 317 10 376 12 421 14 456

ID: 549079 • Letter: W

Question

water displaced (mL)

Time (seconds)

2

174

4

218

6

273

8

317

10

376

12

421

14

456

16

502

18

584

20

614

22

679

water displaced (mL)

Time (seconds)

2

201

4

246

6

292

8

331

10

397

12

436

14

489

16

541

18

584

20

631

22

702

water displaced (mL)

Time (seconds)

2

194

4

229

6

285

8

328

10

382

12

413

14

469

16

512

18

590

20

622

22

691

(a) Assuming the undiluted catalyst solution is 1% mass/volume I3K, how many moles of I3K are present in reactions 1 and 3? The molecular weight of I3K is 420 g/mol.

(b) What is [I3K] present in reactions 1 and 3 in units of molarity?

(c) What is [I3K] present in reaction 2 in units of molarity?

Chart 1: 10 mL undiluted 1-2% IKI; and 5 ml 3% H2O2

Chart 2: 10 mL 0.5-1.0% IKI; and 5 ml 3% H2O2

Chart 3: 10 mL undiluted 1-2% IKI; and 5 ml 2.25% H2O2

water displaced (mL)

Time (seconds)

2

174

4

218

6

273

8

317

10

376

12

421

14

456

16

502

18

584

20

614

22

679

Explanation / Answer

(a) Given, the undiluted catalyst solution is 1% mass/volume I3K

As we know the 1% mass/volume means that 1g of solute in 100 ml of solution.

Volume of undiluted solution = 10 mL

100 ml solution contains I3K catalyst = 1 g

10 ml solution contains I3K catalyst = 1x10/100 = 0.1 g

molecular weight of I3K = 420 g/mol

No. of moles of I3K = weight or mass of I3K/molecular weight of I3K = 0.1g /420 g/mol = 2.381x10-4

Hence, No. of moles of I3K present in the reaction 1 and 3 are 2.381x10-4.

(b) Molarity of the I3K = no. of moles of I3K/volume of solution (L)

We know the no. of moles of the I3K (2.381x10-4 moles) and volume of solution (10 mL + 5 mL = 15 mL).

Hence,

concentration of I3K = [I3K] = (2.381x10-4 / 15 ml)x1000

= 1.587x10-2 M

Hence [I3K] present in reaction 1 and 3 is 2.381x10-2 M

(c) In case of reaction 2, the percentage of I3K become half ( 0.5% mass/volume).

Hence concentration of I3K, in this case, becomes half of the reaction 1 and 3.

[I3K] = 1.587x10-2 M/2 = 7.94x10-3 M

Hence [I3K] present in reaction 2 is 7.94x10-3 M