Chemistry 2024 Jamb Examination
Chemistry 2024 Jamb Examination
Question 1
At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that
According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass:\[ \dfrac{r_X}{r_Y} = \sqrt{\dfrac{M_Y}{M_X}} \]Given that \( r_X = 2 r_Y \), we have:\[ 2 = \sqrt{\dfrac{M_Y}{M_X}} \quad \Rightarrow \quad M_Y = 4 M_X \]This means gas Y is four times as heavy as gas X.
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Question 2
The number of molecules of helium gas contained in 11.5g of the gas is
First, calculate the number of moles of helium:
n = mass / molar mass = 11.5 / 4 = 2.875 moles
The number of molecules is given by Avogadro's number:
N = n × NA = 2.875 × 6.022 × 1023 ≈ 1.73 × 1024
n = mass / molar mass = 11.5 / 4 = 2.875 moles
The number of molecules is given by Avogadro's number:
N = n × NA = 2.875 × 6.022 × 1023 ≈ 1.73 × 1024
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Question 3
The constituents of Alnico are Aluminium, Nickel and
Alnico is an alloy primarily composed of Aluminium (Al), Nickel (Ni), and Cobalt (Co).
It is widely used in the production of permanent magnets due to its high magnetic strength.
It is widely used in the production of permanent magnets due to its high magnetic strength.
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Question 4
Silver and Gold are metals found as free uncombined elements in the earth's crust, which makes them to be referred to as
Metals that are found in nature in their free, uncombined state, resisting corrosion and oxidation, are referred to as noble metals.
Silver (Ag) and Gold (Au) are classic examples of noble metals due to their low reactivity.
Silver (Ag) and Gold (Au) are classic examples of noble metals due to their low reactivity.
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Question 5
In the extraction of Aluminium, the silica impurity is removed by
During the extraction of Aluminium from bauxite, impurities like silica ($SiO_2$) are removed by reacting with sodium hydroxide to form soluble sodium silicate, which can be separated.
This is effectively a precipitation or chemical separation process.
This is effectively a precipitation or chemical separation process.
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Question 6
When Calcium ethynide is decomposed by water, the gas produced is
Calcium ethynide (CaC2) reacts with water to produce acetylene (C2H2) gas and calcium hydroxide:\[ CaC_2 + 2 H_2O \rightarrow C_2H_2 + Ca(OH)_2 \]This is a standard laboratory method for producing acetylene gas.
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Question 7
The constituent of petroleum fraction used in surfacing road is
Bitumen is the heavy, viscous fraction of petroleum used in road surfacing due to its adhesive and waterproof properties.
Other fractions like petrol, diesel, and lubricating oil are used as fuels or lubricants.
Other fractions like petrol, diesel, and lubricating oil are used as fuels or lubricants.
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Question 8
During the fractional distillation of crude oil, the fraction that distills at 200 - 250°C is
Kerosene is the petroleum fraction that distills between 200°C and 250°C.
It is commonly used as fuel for jet engines, heating, and lamps.
It is commonly used as fuel for jet engines, heating, and lamps.
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Question 9
A liquid hydrocarbon obtained from fractional distillation of coal tar that is used in the pharmaceutical industry is
Benzene is a liquid hydrocarbon obtained from the fractional distillation of coal tar and is widely used in the pharmaceutical and chemical industries as a solvent and starting material for organic synthesis.
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Question 10
An organic compound contains 53.1% Carbon, 6.2% Hydrogen, 12.4% Nitrogen, and 28.3% Oxygen by mass. What is the molecular formula of the compound if its vapour density is 56.5? [C =12, H = 1, N = 14, O = 16].
Calculate the empirical formula:
Number of moles of each element per 100g:
C: 53.1/12 = 4.425 ≈ 4.43
H: 6.2/1 = 6.2
N: 12.4/14 = 0.885 ≈ 0.89
O: 28.3/16 = 1.77 ≈ 1.77
Divide by smallest (0.885):
C: 4.43/0.885 ≈ 5
H: 6.2/0.885 ≈ 7
N: 0.885/0.885 = 1
O: 1.77/0.885 ≈ 2
Empirical formula = $C_5H_7NO_2$.
Vapour density confirms the molecular formula is the same as empirical, hence $C_5H_7NO_2$..
Number of moles of each element per 100g:
C: 53.1/12 = 4.425 ≈ 4.43
H: 6.2/1 = 6.2
N: 12.4/14 = 0.885 ≈ 0.89
O: 28.3/16 = 1.77 ≈ 1.77
Divide by smallest (0.885):
C: 4.43/0.885 ≈ 5
H: 6.2/0.885 ≈ 7
N: 0.885/0.885 = 1
O: 1.77/0.885 ≈ 2
Empirical formula = $C_5H_7NO_2$.
Vapour density confirms the molecular formula is the same as empirical, hence $C_5H_7NO_2$..
