Chemistry 2014 Jamb Examination
Chemistry 2014 Jamb Examination
Question 41
The radioactive radiation used in studying the arrangement of particles in giant organic molecules is
X-rays possess wavelengths that are comparable to interatomic distances in solids and large molecules, allowing them to be effectively used in determining atomic arrangements. When directed at a crystalline or complex organic structure, X-rays are diffracted, and the resulting diffraction pattern provides information about the spatial arrangement of atoms. This principle underlies X-ray crystallography, a key technique used to study the structure of proteins, DNA, and other macromolecules.
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Question 42
A silicon-containing ore has 92% 28Si, 5% 29Si and 3% 30Si. Calculate the relative atomic mass of the silicon.
The relative atomic mass of silicon is obtained by taking the weighted average of its isotopes.\[ R.A.M = \]\[\dfrac{(92 \times 28) + (5 \times 29) + (3 \times 30)}{100} \]\[ = \dfrac{2576 + 145 + 90}{100} \]\[ = \dfrac{2811}{100} = 28.11 \]This value represents the average atomic mass of naturally occurring silicon, accounting for the isotopic abundances.
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Question 43
A mixture is different from a compound because
In a compound, atoms of the constituent elements are chemically combined in a fixed ratio through chemical bonds, forming a new substance with properties distinct from its elements. In contrast, a mixture is a physical combination of substances without chemical bonding, and the components retain their individual properties. Examples include air (mixture) and water (compound).
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Question 44
What is the percentage of sulphur in sulphur (IV) oxide?
To find the percentage of sulphur in SO2:
Determine molar mass.\[ M = 32 + (16 \times 2) = 64 \; \text{g mol}^{-1} \] Calculate sulphur's contribution.\[ \%S = \dfrac{32}{64} \times 100 = 50\% \]This means sulphur constitutes half the total mass of sulphur (IV) oxide.
Determine molar mass.\[ M = 32 + (16 \times 2) = 64 \; \text{g mol}^{-1} \] Calculate sulphur's contribution.\[ \%S = \dfrac{32}{64} \times 100 = 50\% \]This means sulphur constitutes half the total mass of sulphur (IV) oxide.
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Question 45
A gas X diffuses twice as fast as gas Y. If the relative molecular mass of X is 32, calculate the relative molecular mass of Y.
Using Graham’s law of diffusion:\[ \dfrac{r_X}{r_Y} = \sqrt{\dfrac{M_Y}{M_X}} \]Given \( r_X / r_Y = 2 \) and \( M_X = 32 \):
\[ 2 = \sqrt{\dfrac{M_Y}{32}} \]Squaring both sides:\[ 4 = \dfrac{M_Y}{32} \]\[ M_Y = 128 \]Hence, the relative molecular mass of gas Y is 128, indicating it diffuses more slowly because it is heavier.
\[ 2 = \sqrt{\dfrac{M_Y}{32}} \]Squaring both sides:\[ 4 = \dfrac{M_Y}{32} \]\[ M_Y = 128 \]Hence, the relative molecular mass of gas Y is 128, indicating it diffuses more slowly because it is heavier.
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Question 46
200 cm3 of a gas at 25°C exerts a pressure of 700 mmHg. Calculate its pressure if its volume increases to 350 cm3 at 75°C.
Using the combined gas law:\[ \dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2} \]
Given: P1 = 700 mmHg, V1 = 200 cm3, T1 = 25 + 273 = 298 K V2 = 350 cm3, T2 = 75 + 273 = 348 K
Substitute into the equation:\[ P_2 = \dfrac{P_1 V_1 T_2}{T_1 V_2} \]\[ P_2 = \dfrac{700 \times 200 \times 348}{298 \times 350} \]\[ P_2 = 467.11 \; \text{mmHg} \]The pressure decreases due to the increase in volume despite the temperature rise.
Substitute into the equation:\[ P_2 = \dfrac{P_1 V_1 T_2}{T_1 V_2} \]\[ P_2 = \dfrac{700 \times 200 \times 348}{298 \times 350} \]\[ P_2 = 467.11 \; \text{mmHg} \]The pressure decreases due to the increase in volume despite the temperature rise.
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Question 47
An element X has electron configuration 1s2 2s2 2p6 3s2 3p5. Which of the following statements is correct about the element?
The outermost shell (n = 3) has the configuration 3s2 3p5, meaning there are seven valence electrons. Elements with seven valence electrons belong to Group 17, known as the halogens. They readily gain one electron to form stable octets and exist naturally as diatomic molecules, such as chlorine (Cl2).
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Question 48
The general formula of an alkyl halide (where X represents the halide) is
An alkyl halide, also called a haloalkane, is derived from an alkane by replacing one hydrogen atom with a halogen atom (F, Cl, Br, or I). The general formula of alkanes is \( C_nH_{2n+2} \), so removing one hydrogen gives the alkyl group \( C_nH_{2n+1} \). Substituting one halogen yields the general formula \( C_nH_{2n+1}X \).

