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post-utme2017chemistry
Q1. The reddish-brown rust on iron roofing sheets consists of
AFe³⁺(H₂O)₆
BFeO·H₂O
CFe₂O₃·XH₂O
DFe₃O₄·2H₂O
Correct Answer: C
Rust is hydrated iron (III) oxide, represented as Fe₂O₃·XH₂O (where X is variable). It is formed by the oxidation of iron in the presence of water and oxygen. The reddish-brown colour is characteristic of Fe₂O₃ (iron III oxide) in its hydrated form.
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post-utme2017chemistry
Q2. Incomplete oxidation of ethanol yields
ACH₃COOH
BCH₃COCH₃
CCH₃CH₂OCH₂CH₃
DCH₃CHO
Correct Answer: D
Ethanol (CH₃CH₂OH) undergoes stepwise oxidation. Incomplete (mild) oxidation yields ethanal (CH₃CHO), an aldehyde. Complete (vigorous) oxidation gives ethanoic acid (CH₃COOH). The incomplete oxidation stops at the aldehyde stage.
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post-utme2017chemistry
Q3. The diagram above. Y is
Afused CaO
BH₂O
CNaOH
DConcentrated H₂SO₄
Correct Answer: D
In the diagram, Y is the conical flask containing the drying agent through which the gas passes before collection. Concentrated H₂SO₄ is the standard drying agent used to dry acidic or neutral gases in laboratory gas preparation. Z is the thistle funnel for adding acid, and X is the water trough for collection by downward displacement of water.
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post-utme2017chemistry
Q4. The salt formed from a weak acid and a strong base hydrolyzes in water to form
AA saturated solution
Ban acidic solution
Ca buffer solution
Dan alkaline solution
Correct Answer: D
A salt formed from a weak acid and a strong base (e.g., CH₃COONa from acetic acid + NaOH) undergoes hydrolysis in water. The anion of the weak acid accepts a proton from water, producing OH⁻ ions. This makes the solution alkaline (pH > 7).
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post-utme2017chemistry
Q5. Tin is unaffected by air at ordinary temperature due to its
ALow melting point
BWeak electropositive character
CHigh boiling point
DWhite lustrous appearance
Correct Answer: B
Tin is resistant to corrosion at ordinary temperatures because of its weak electropositive (low reactivity) character. It has a low tendency to lose electrons and oxidize in air. This weak electropositive character means it does not readily react with oxygen or moisture at room temperature, making it useful as a protective coating (tinning).
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post-utme2017chemistry
Q6. The shape of ammonia molecules is
Atrigonal pyramidal
Boctahedral
Csquare planar
Dtetrahedral
Correct Answer: A
NH₃ has 3 bonding pairs and 1 lone pair of electrons around the nitrogen. The lone pair repels the bonding pairs more strongly, compressing the H-N-H bond angle to ~107°. This gives a trigonal pyramidal shape (not tetrahedral, which is the electron geometry).
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post-utme2017chemistry
Q7. Due to the high reactivity of sodium, it is usually stored under
Awater
Bmercury
Cparaffin
Dphenol
Correct Answer: C
Sodium is stored under paraffin oil (kerosene/mineral oil) to prevent it from reacting with oxygen and moisture in the air. Paraffin is a non-polar, unreactive liquid that forms a protective layer around the sodium. Water would react explosively with sodium; mercury would form an amalgam; phenol would react with sodium.
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post-utme2017chemistry
Q8. An oxide XO₂ has a vapour density of 32. What is the atomic mass of X?
A20
B32
C14
D12
Correct Answer: B
Molar mass of XO₂ = 2 × vapour density = 2 × 32 = 64 g mol⁻¹. Molar mass = atomic mass of X + 2(16) → X + 32 = 64 → X = 32. The element with atomic mass 32 is sulphur (S), and XO₂ is SO₂ (sulphur dioxide).
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post-utme2017chemistry
Q9. If 100 cm³ of oxygen passes through a porous plug in 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]
A10.0 s
B12.5 s
C17.7 s
D32.0 s
Correct Answer: B
By Graham's Law, rate ∝ 1/√M and rate ∝ 1/time. Therefore: t(H₂)/t(O₂) = √(M(H₂)/M(O₂)) = √(2/32) = √(1/16) = 1/4. t(H₂) = 50/4 = 12.5 s. H₂ is lighter and diffuses faster, taking less time.
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post-utme2017chemistry
Q10. Calculate the amount in moles of a gas which occupies 10.5 dm³ at 6 atm and 30°C [R = 0.082 atm dm³ K⁻¹ mol⁻¹]
A2.536
B1.623
C4.736
D0.394
Correct Answer: A
Using the ideal gas equation PV = nRT. T = 30 + 273 = 303 K. n = PV/RT = (6 × 10.5)/(0.082 × 303) = 63/24.846 = 2.536 mol.
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post-utme2017chemistry
Q11. The alkyl group is represented by the general formula
ACₙH₂ₙ
BCₙH₂ₙ - 2
CCₙH₂ₙ + 1
DCₙH₂ₙ + 2
Correct Answer: C
An alkyl group is derived from an alkane by removing one hydrogen atom. Alkanes have the formula CₙH₂ₙ₊₂, so removing one H gives the alkyl group CₙH₂ₙ₊₁. Examples: methyl (CH₃, n=1), ethyl (C₂H₅, n=2), propyl (C₃H₇, n=3).