Chemistry 2007 Jamb Examination

Chemistry 2007 Jamb Examination
Available Sections
2. Career
4. Job
12345

Question 21

Calculate the number of moles of HCl present in 20 cm³ of a 0.75 M solution of the acid
Show explanation Expand Question

Number of moles (n) is given by:

\[ n = C \times V \]

Where C = concentration in mol dm⁻³, V = volume in dm³.

Convert 20 cm³ to dm³: \( 20/1000 = 0.020\,\text{dm³} \).

Substitute values: \( n = 0.75 \times 0.020 = 0.015\,\text{mol} \).

Hence, the number of moles of HCl is 0.015 mol (option A).

Not clear enough?

Question 22

The oxidation state of oxygen in tetraoxosulphate (VI) acid is
Show explanation Expand Question

Tetraoxosulphate (VI) acid = H₂SO₄.

Let the oxidation state of S be x, O is usually -2. Total charge = 0:

\[ 2(+1) + x + 4(-2) = 0 \]

\[ 2 + x - 8 = 0 \]

\[ x = +6 \]

Thus, the oxidation state of oxygen = -2 (option B).

Not clear enough?

Question 23

MnO₂(s) + x HCl(aq) → MnCl₂(aq) + y H₂O(l) + z Cl₂(g). Determine the values of x, y, z respectively

Show explanation Expand Question
Write the balanced redox reaction:
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
Count the molecules:
x = 4 (HCl)
y = 2 (H2O)
z = 1 (Cl2)
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
Therefore: x = 4, y = 2, z = 1
Not clear enough?

Question 24

Calculate the mass of chlorine gas which occupies a volume of 1.12 dm³ at s.t.p [Cl = 35.5, molar volume = 22.4 dm³]

Show explanation Expand Question

Calculate moles using molar volume:

\[ n = V/V_m = 1.12/22.4 = 0.05\,\text{mol} \]

Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol

Mass = n × M = 0.05 × 71 = 3.55 g

Oops! Option C = 3.55 g. Correct answer is 3.55 g (option C).

Not clear enough?

Question 25

Which experiment led to the measurement of the charge on an electron?
Show explanation Expand Question

The Millikan oil-drop experiment measured the charge of a single electron by observing the motion of charged oil droplets in an electric field. Option B is correct.

Not clear enough?

Question 26

What is the volume of oxygen required to burn completely 45 cm³ of methane at s.t.p?

Show explanation Expand Question

The balanced combustion reaction of methane (CH₄) is:

\[ CH_4 + 2O_2 → CO_2 + 2H_2O \]

This shows that 1 volume of CH₄ reacts with 2 volumes of O₂.

Given volume of methane = 45 cm³

\[ V_{O_2} = 2 imes 45 = 90 \text{ cm³} \]

Correction: Actually option D = 90.0 cm³. Hence the oxygen volume required is 90 cm³ (option D).

Not clear enough?

Question 27

How many electrons are in the L shell of ³¹P?
Show explanation Expand Question

Atomic number of phosphorus = 15

Electron configuration: 1s² 2s² 2p⁶ 3s² 3p³

L-shell = 2nd shell = 2s² + 2p⁶ = 8 electrons

Hence, number of electrons in the L-shell = 8 (option B).

Not clear enough?

Question 28

If the electron configuration of an element is 1S2 2S2 2P5, how many unpaired electrons are there?

Show explanation Expand Question

Electron configuration: 1s² 2s² 2p⁵

2p orbital has 3 sub-orbitals: ↑↓, ↑, ↑ (one unpaired electron)

Hence, there is 1 unpaired electron (option C).

Not clear enough?

Question 29

Which of the following decreases when a given mass of gas is compressed to half its initial volume?
Show explanation Expand Question

Compressing a gas reduces its volume.

Average intermolecular distance \( r \) is proportional to \( V^{1/3} \). Halving volume decreases \( r \).

Other quantities:

  • Atomic radius does not change.
  • Number of molecules remains constant.
  • Frequency of collisions increases (does not decrease).

Hence, the average intermolecular distance decreases (option B).

Not clear enough?

Question 30

The elements that belong to the third period of the periodic table are?
Show explanation Expand Question

The third period of the periodic table contains elements with principal quantum number n=3: Na, Mg, Al, Si, P, S, Cl, Ar.

Option D includes: Na, Mg, S, Ar, which are all in the third period.

Hence correct answer is option D.

Not clear enough?

12345
In this page
Chemistry 2007 Jamb Examination
Open Model!