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

Answer 1

(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