HIGHER CHEMISTRY REVISION.

Unit 1:- The Mole

Using information from the above graph, calculate the mass of hydrogen peroxide used in the reaction, assuming all the hydrogen peroxide

decomposed.

(Take the molar volume of oxygen to be 24 litres mol-1)

1. The balanced equation for the decomposition of hydrogen peroxide into water and oxygen is:

2H2O2 (l) 2H2O(l) + O2 (g)

24000 cm3 of gas = 1 mol of O2

So 40 cm3 = 0.0167 mol

1 mole of O2 weighs 32g

So 0.0167 mol weighs 0.534g

2. . A student electrolysed dilute sulphuric acid using the apparatus shown in

order to estimate the volume of one mole of hydrogen gas.

Current = 0.5 A

Time = 14 minutes.

Volume of hydrogen collected = 52 cm3.

(a) Calculate the molar volume of hydrogen gas.

(b) What change could be made to the apparatus to reduce a possible

significant source of error?

(a) Q= I x t

= 0.5 x 14 x 60

= 420C

2H+ + 2e - H2

2 x 96500C 1 mol

193000C 1 mol

420C gives 52 cm3

So 193000C gives 23 895 cm3

or 23.895 litres

(b) Introduce a variable resistor

so that current can be kept

constant.

3. The symbol for the Avogadro Constant is L.

Identify the two true statements.

A 64.2 g of sulphur contains approximately L atoms.

B 16.0 g of oxygen contains approximately L molecules.

C 6.0 g of water contains approximately L atoms.

D 1.0g of hydrogen contains approximately L protons.

E 2.0 litres of 0.50 mol l-1 sulphuric acid contains approximately

L hydrogen atoms.

F 1.0 litres of 1.0 mol l -1 barium hydroxide solution contains

approximately L hydroxide atoms.

C and D

4. A student heated a compound which gave off carbon dioxide

gas and water vapour.

The volume of carbon dioxide collected was 240 cm3.

Calculate the number of molecules in this volume.

(Take the molar volume of carbon dioxide to be 24 litres mol-1.)24 litres = 24000 cm3.

24000 cm3 contains 6.02 x 1023 CO2 molecules.

So 240 cm3 contains 240/24000 x 6.02 x 1023 = 6.02 x 1021 CO2 molecules.

A student added 0.20 g of silver nitrate, AgNO3, to 25 cm3 of water. This solution was then added to 20 cm3 of 0.0010 mol l-1 hydrochloric acid as shown in the diagram

The equation for the reaction which occurs is:

AgNO3 (aq) + HCl(aq) AgCl(s) + HNO3 (aq)

(a) Name the type of reaction which takes place.

(b) Show by calculation which reactant is in excess.

(a) Precipitation.

(b) No of moles of HCl = C x V(litres) = 20/1000 x 0.001

= 2 x 10-5 mol

No of moles of AgNO3 = mass/gfm = 0.2/169.9

= 1.2 x 10-3 mol

So the silver nitrate is in excess.

5.

6. When sodium hydrogencarbonate is heated to 112oC it decomposes and the gas

carbon dioxide is given off:

2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)

The following apparatus can be used to measure the volume of carbon dioxide

produced in the reaction.

(a) Why is an oil bath used and not a water bath?

(b) (i) Calculate the theoretical volume of carbon dioxide produced by the complete

decomposition of 1.68 g of sodium hydrogencarbonate.

Take the molar volume of carbon dioxide to be 23 litres mol-1

(ii) Assuming that all of the sodium hydrogen carbonate is decomposed, suggest

why the volume of carbon dioxide collected in the measuring cylinder would

be less than the theoretical value.

(a) Water boils at 100oC and

so could not raise the

temperature to 112oC.

(b) (i) From equation

2mol of NaHCO3

gives 1 mol of CO2.

i.e. 168 g NaHCO3 44g CO2.

So 1.68 g 0.44g

b) (ii) Some of the carbon dioxide

dissolves in water.

7. The concentration of a solution of sodium thiosulphate can be found by reaction with iodine.

The iodine is produced by electrolysis of an iodide solution using the apparatus shown.

The current is noted and the time when the indicator detects the end-point of the

reaction is recorded.

Iodine is produced from the iodide solution according to the following equation:

2I-(aq) I2(aq) + 2e-

(a) Calculate the number of moles of iodine generated during the electrolysis given

the following results.

Current = 0.010 A Time = 1 min 37 sQ = I x t = 0.01 x 97 = 0.97C

2I-(aq) I2(aq) + 2e-

1 mol 2F

1 mol 2 x 96500 C

1 mol 193000C

So 0.97 C 0.97/193000 x 1 = 5 x 10-6 mol

7. (b) The iodine produced reacts with the thiosulphate ions according to the equation:

I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

iodine thiosulphate ions

At the end-point of the reaction, excess iodine is detected by the indicator.

(i) Name the indicator which could be used to detect the excess iodine present

at the end-point.

(ii) In a second experiment it was found that 1.2 x 10-5 mol of iodine reacted with

3.0 cm3 of the sodium thiosulphate solution.

Use this information to calculate the concentration of the sodium thiosulphate

solution in mol l-1.(b) (i) Starch solution.

(ii) I2(aq) + 2S2O32-(aq) 2I-(aq) + S4O6

2-(aq)

1 mol 2 mol

So 1.2 x 10-5 mol of I2(aq) reacts with 2.4 x 10-5 mol of S2O32-(aq

Concentration of thiosulphate = no of moles/volume (litres)

=

= 0.008 mol l-1.

2.4 x 10-5

0.003

8. Aluminium is manufactured in cells by the electrolysis of aluminium oxide

dissolved in molten cryolite.

What mass of aluminium is produced each hour, if the current passing

through the liquid is 180 000 A?Al3+(l) + 3e- Al(l)

3 F 1 mol

3 x 96500C 27 g

289500C 27 g

Q = I x t

= 180 000 x 60 x 60

= 648 000 000 C

289500C 27 g

So 648 000 000 C 648 000 000/289500 x 27

= 60 435 g

= 60.435 kg of Al

9. Calcite is a very pure form of calcium carbonate which reacts with nitric acid as follows:

CaCO3(s) + 2HNO3 (aq) Ca(NO3) 2(aq) + H2O(l) + CO2(g)

A 2.14 g piece of calcite was added to 50.0 cm3 of 0.200 mol l-1 nitric acid in a beaker.(a) Calculate the mass of calcite, in grams, left unreacted.(b) Describe what could be done to check the result obtained in (a)(a) No. of moles of calcite = mass/gfm = 2.14/100 = 0.0214 mol.

No. of moles of acid = C x V(litres) = 0.2 x 50/1000 = 0.01

From equation 0.0214 mol of calcite reacts with 0.0418 mol of acid.

So all the acid is used up

0.01 mol of acid reacts with 0.005 mol of calcite = 0.005 x 100 = 0.5 g

Mass of calcite left over = 2.14 – 0.5g = 1.64 g.

(b) Filter off any unreacted calcite, dry and then weigh it.

10. Diphosphine, P2H4, is a hydride of phosphorus. All of the covalent bonds

in diphosphine molecules are non-polar because the elements present have

the same electronegativity.

(a) What is meant by the term”electronegativity”?

(b) The balanced equation for the complete combustion of diphosphine is:

2P2H4(g) + 7O2 (g) P4O10(s) + 4H2O(l)

What volume of oxygen would be required for the complete combustion o

10 cm3 of diphosphine?

(c) Calculate the volume occupied by 0.330 g of diphosphine.

(Take the molar volume to be 24.0 litres mol-1.)

(a) Electronegativity is a measure of the attraction of an atom

for electrons it shares with other atoms.

(b) 2P2H4(g) + 7O2 (g) P4O10(s) + 4H2O(l)

2 mol 7 mol

2 vol 7 vol

10cm3 35cm3

(c) 1 mole of P2H4 weighs 66g

66g occupies 24.0 litres

So 0.330g occupies 0.33/66 x 24 = 0.12 litres

11. In 1996, the scientists Robert Curl, Harold Kroto and Richard Smalley won the

Nobel Prize in Chemistry for their contribution to the discovery of new forms of

carbon called fullerines.

(a) In what way does the structure of fullerines differ from the other forms of

carbon, diamond and graphite?

(b) One form of fullerine, C60, forms a superconducting crystalline compound with

potassium.

Its formula can be represented as K3C60.

A sample of this compound was found to contain 2.88 g of carbon.

(i) Calculate the number of moles of fullerine used to make this compound.

(ii) Calculate the mass of potassium, in grams, in the sample.

(a) Fullerines are covalent molecular solids – diamond and graphite are covalen

network solids.

(b) 1 mole of K3C60 contains 60 x 12 = 720g of carbon.

So 2.88g of carbon = 2.88/720 = 0.004 mol of fullerine.

(c) 1 mole of K3C60 contains 3 x 39 g of potassium.

So 0.004 moles contains 3 x 39 x 0.004 g

= 0.468 g of potassium

12. Ionisation energies provide information about the structure of atoms.

(a) Write the equation, showing state symbols, for the first ionisation

energy of sodium.

(b) Calculate the number of electrons lost when one mole of boron atoms

is converted into one mole of boron atoms with a charge of 3+.

(a) Na(g) Na+(g) + e-

(b) B(g) B3+(g) + 3e-

1 mol 3 mol

3 x 6.02 x 1023

1.806 x 1024 electrons

13. The ion-electron equation for the production of zinc in an

electrolysis cell is

Zn2+ + 2e- Zn

If a current of 2000 A is used in the cell, calculate the mass of

zinc, in kg, produced in 24 hours.

Zn2+ + 2e- Zn Q= I x t

2 F 1 mol = 2000 x x 24 x60 x 60

193000C 65.4 g = 172.8 x 106C

So 172.8 x 106 C 58555g or 58.55 kg