WAEC Chemistry Past Questions & Answers - Page 4

16.

Positive ions in a sea of electrons are found in

A.

covalent bonds

B.

metallic bonds

C.

dative bonds

D.

ionic bonds

Correct answer is B

In metallic bonding, positive metal ions (cations) are surrounded by a "sea" of delocalized electrons that move freely throughout the metal lattice. This arrangement creates a cohesive force that holds the metal atoms together. The mobile electrons contribute to the unique properties of metals, such as high electrical and thermal conductivity, malleability, and ductility.

17.

Determine the volume of 0.100 mol of HCl in \(0.250 mol dm^3\) of solution.

A.

\(200cm^3\)

B.

\(100cm^3\)

C.

\(20cm^3\)

D.

\(400cm^3\)

Correct answer is D

Molarity = 0.100 mol / 0.250 mol dm3
= 0.100 mol / 0.250 dm3 * 1000 cm3/dm3
= 400 cm3
Volume (V) = Amount (n) / Concentration (C)
Given that the amount of HCl is 0.100 mol and the concentration of the solution is 0.250 mol/dm³, you can calculate the volume:
Volume (V) = 0.100 mol / 0.250 mol/dm³ = 0.4 dm³
Since 1 dm³ is equal to 1000 cm³, you can convert the volume from dm³ to cm³:
Volume (V) = 0.4 dm³ × 1000 cm³/dm³ = 400 cm³
Therefore, the volume of 0.100 mol of HCl in a 0.250 mol/dm³ solution is 400 cm³.

18.

Which of the statements about gases is not correct?

A.

Gases have low densities compared to solids and liquids of equal mass.

B.

The total kinetic energy of the gas is not affected by collision

C.

Molecules of the gas are in constant motion

D.

Gases are highly soluble in water at high temperatures

Correct answer is D

The statement that is not correct about gases is Gases are highly soluble in water at high temperatures.
Gases are not highly soluble in water at high temperatures. The solubility of gases in water decreases as the temperature increases. This is because the kinetic energy of the gas molecules increases as the temperature increases. The increased kinetic energy of the gas molecules makes it more difficult for them to be dissolved in water.

19.

A compound that could be dried by using conc. tetraoxosulphate (VI) acid and not by calcium oxide is likely to be

A.

a deliquescent salt

B.

a neutral salt

C.

an alkaline gas

D.

an acid anhydride

Correct answer is D

The compound that could be dried by using conc. tetraoxosulphate (VI) acid and not calcium oxide is an acid anhydride.
Acid anhydrides are compounds that can react with water to form acids. When acid anhydrides are exposed to air, they will absorb water from the air and become hydrated. This can make them difficult to dry with calcium oxide, which is a base.

20.

If \(100^3\) of a saturated solution of sodium tetraoxosulphate (VI) at 30° C contains 10.5 g of the salt, what would be its solubility at this temperature? [(\(Na_2SO_4\) = 142]

A.

2.15 mol dm-3

B.

0.74 mol dm-3

C.

0.60 mol dm-3

D.

0.57 mol dm-3

Correct answer is B

Molar mass of Na2SO4 = 2(Na) + S + 4(O) = 2(23.0 g/mol) + 32.1 g/mol + 4(16.0 g/mol) = 46.0 g/mol + 32.1 g/mol + 64.0 g/mol = 142.1 g/mol

Now, we can calculate the number of moles of Na2SO4 in the solution:

Number of moles = Mass of the salt / Molar mass
Number of moles = 10.5 g / 142.1 g/mol ≈ 0.0739 mol

Next, we need to find the volume of the saturated solution, given as 100 cm³, and convert it to liters:

Volume of solution = 100 cm³ = 100 cm³ / 1000 cm³/dm³ = 0.1 dm³

Finally, we can calculate the solubility in mol/dm³:
Solubility = Number of moles / Volume of solution
Solubility = 0.0739 mol / 0.1 dm³ ≈ 0.739 mol/dm³

Rounded to two decimal places, the solubility of sodium tetraoxosulphate (VI) at 30°C is approximately 0.74 mol/dm³.
Therefore, the correct option is:
0.74 mol dm-3.
Solubility = 10.5 g / 100 cm3 * 142 g/mol = 0.739 mol/dm3