Chemistry 2006 Jamb Examination

Chemistry 2006 Jamb Examination
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Question 21

When 50cm3 of a saturated solution of sugar of molar mass 342 g at 40oC was evaporated to dryness, 34.2g of dry solid was obtained. The solubility of sugar at 40oC is

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Calculation (showing steps):

One sample: 50 cm3 = 0.050 dm3 contains 34.2 g sugar.
Molar mass M = 342 g mol-1.
Concentration g·dm-3\[=\dfrac{34.2\ \text{g}}{0.050\ \text{dm}^3}=684\ \text{g·dm}^{-3} \] molarity (mol·dm-3)\[=\dfrac{684\ \text{g·dm}^{-3}}{342\ \text{g·mol}^{-1}}\]\[=2.00\ \text{mol·dm}^{-3} \]

Therefore the solubility is 2.0 mol·dm-3. This matches option A.

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Question 22

Which of the following results in the fall of acid rain?
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Cause: Acid rain is primarily caused by atmospheric oxidation of sulphur dioxide (SO₂) and nitrogen oxides (NOₓ) to form strong acids (sulphuric and nitric acids) which dissolve in rainwater.

Key reactions:
2SO₂ (g) + O₂ (g) → 2SO₃ (g)
SO₃ (g) + H₂O (l) → H₂SO₄ (aq)

These acids lower the pH of precipitation. The second paragraph: oxides of sulphur are the main contributors (option C). Other pollutants can worsen effects, but SO₂/NOₓ are the primary drivers of acid rain.

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Question 23

Water has the ability to dissolve almost all substances because its molecules has a structure with
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Polarity: A water molecule (H–O–H) is polar: the oxygen atom is more electronegative and pulls electron density toward itself, producing a partial negative charge at the O-end and partial positive charges on the H-ends. In effect the molecule has a positive side and a negative side.

This dipole (one positive end and one negative end) makes water an excellent solvent for ionic and polar substances because oppositely charged parts of the water molecule stabilize ions and polar molecules in solution. Therefore option B is correct.

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Question 24

When a crystal was added to the clear solution of its salt, the crystal did not dissolve and the solution remained unchanged. This showed that the solution was
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If adding a seed crystal does not dissolve and the solution remains unchanged, the solution is at equilibrium with the solid phase — i.e. it is saturated. In a saturated solution the maximum amount of solute at that temperature is already dissolved.

In contrast, a supersaturated solution would cause the crystal to induce rapid crystallization (the crystal would grow), while an unsaturated solution would dissolve the added crystal. Thus the correct answer is saturated (D).

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Question 25

A particle that contains 11 protons, 12 neutrons and 10 electrons is probably a
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Count: 11 protons → atomic number 11 (sodium, Na). Neutral Na atom would have 11 electrons, but this particle has 10 electrons, so it carries a net +1 charge (Na⁺). Na⁺ is a cation — a metallic ion.

Therefore option B (metallic ion) is correct. The second paragraph explains the logic: proton number identifies the element; loss of one electron yields the common Na⁺ ion.

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Question 26

A difference between chemical and physical change is that in a chemical change
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Chemical change produces one or more new substances with different chemical properties and compositions. For example, hydrogen and oxygen react to form water:

\[2H_2 (g) + O_2 (g) \rightarrow 2H_2O (l)\]

This reaction yields a substance (H2O) that has properties different from the reactants.

Physical changes (melting, evaporation, dissolution without reaction) do not form new substances; they only change state or form. Therefore the hallmark difference is that a new substance is formed (option D).

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Question 27

The volume of 2.5 moles of oxygen at s.t.p is [Molar volume of a gas at s.t.p = 22.4 dm3]
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At STP one mole of any ideal gas occupies 22.4 dm3. For 2.5 moles:

\[V = n \times 22.4\ \text{dm}^3 = 2.5 \times 22.4 = 56.0\ \text{dm}^3.\]

Thus the required volume is 56.0 dm3 (option D). The first paragraph shows the calculation; the second notes that molar volume at STP is the constant used.

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Question 28

Nuclear fission involves the splitting of a heavy element into two nuclei of
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Nuclear fission is the splitting of a heavy nucleus (e.g., U-235) into two lighter nuclei of roughly comparable mass, along with neutrons and a large amount of energy and radiation. A typical schematic:

\[^{235}\text{U} + n \rightarrow\;^{92}\text{Kr} +\;^{141}\text{Ba} + 3n + \text{energy}\]

This process releases both kinetic energy and prompt radiation; the product masses are nearly equal but not exactly identical. Option A correctly captures this description.

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Question 29

What is the molar mass of a substance, if 0.4 mole of the substance has a mass of 25.0g?
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Molar mass M is mass per mole. Given mass = 25.0 g for n = 0.4 mol,

\[M = \dfrac{\text{mass}}{n}=\dfrac{25.0\ \text{g}}{0.4\ \text{mol}}=62.5\ \text{g·mol}^{-1}.\]

Thus the molar mass is 62.5 g·mol⁻¹ (option A). The first paragraph shows the formula; the second explains that molar mass links mass and amount of substance.

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Question 30

The element with the electron configuration 1s² 2s² 2p³ is

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The configuration 1s² 2s² 2p³ has a total of 2+2+3 = 7 electrons, so atomic number Z = 7. The element with Z = 7 is nitrogen (N). Its ground-state configuration is exactly 1s² 2s² 2p³.

Therefore option D (nitrogen) is correct. The second sentence: count electrons → identify atomic number → match the element.

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Chemistry 2006 Jamb Examination
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