JEE Advanced (Single Correct Type): Some Basic Concepts of Chemistry | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. A pure substance which contains only one type of atom is called________.
(a) An element
(b) A compound
(c) A solid
(d) A liquid

Correct Answer is option (a)
An element is made up of only one type of atom.

Q.2. The smallest particle that can take part in chemical reactions is ————–.
(a) Atom
(b) Molecule
(c) Both (a) and (b)
(d) None of these

Correct Answer is option (c)
The smallest particle that can take part in chemical reactions is both an atom and a molecule.

Q.3. Which of the following is a homogeneous mixture?
(a) Mixture of soil and water
(b) Sugar solution
(c) Mixture of sugar, salt and sand
(d) Iodised table salt

Correct Answer is option (b)
Sugar solution is a homogeneous mixture. A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture

Q.4. The significant figures in 0.00051 are ________.
(a) 5
(b) 3
(c) 2
(d) 26

Correct Answer is option (c)
The significant figures in 0.00051 are 2.

Q.5. Formation of CO and CO₂ illustrates the law of ________.
(a) Law of conservation of mass
(b) Law of Reciprocal proportion
(c) Law of Constant Proportion
(d) Law of Multiple Proportion

Correct Answer is option (d)
If an element forms more than one compound with another element for a given mass of an element, masses of other elements are in the ratio of small whole numbers.

Q.6. The number of significant figures in 6.02 x 1023 is ________.
(a) 23
(b) 3
(c) 4
(d) 26

Correct Answer is option (b)
The number of significant figures in 6.02 x 10²³ is 3

Q.7. The prefix 1018 is ________.
(a) giga
(b) exa
(c) kilo
(d) mega

Correct Answer is option (b)
The prefix 10¹⁸ is exa

Q.8. The mass of an atom of carbon is ————–.
(a) 1g
(b) 1.99 x 10^-23 g
(c) 1/12 g
(d) 1.99 x 10²³ g

Correct Answer is option (b)
The mass of an atom of carbon is {12 / (6.02 x 10²³)} = 1.99 x 10^-23 g

Q.9. A measured temperature on Fahrenheit scale is 200F. What will this reading be on the Celsius Scale?
(a) 40^oC
(b) 94^oC
(c) 93.3^oC
(d) 30^oC

Correct Answer is option (c)
The relationship between Fahrenheit and degree Celsius is: (^oF) = 9/5 (^oC) +32.

Q.10. Which of the following pairs of gases contains the same number of molecules?
(a) 16 g of O₂ and 14 g of N₂ 
(b) 6 g of O₂ and 22 g of CO₂ 
(c) 28 g of N₂ and 22 g of CO₂ 
(d) 32 g of CO₂ and 32g of N₂

Correct Answer is option (a)
Divide the given mass by its molar mass to get moles, then multiply times 6.022×10²³ to get the number of molecules.

Q.11. For a given mixture of NaHCO₃ and Na₂CO₃, volume of a given HCl required is x ml with phenolphthalein indicator and further y ml required with methyl orange indicator. Hence volume of HCl for complete reaction of original NaHCO₃ is
(a) 2x
(b) x/2
(c) y
(d) (y- x)

Correct Answer is option (d)
Using phenolphthalein,
Na₂CO₃ +  HCl  → NaHCO₃ +  NaCl
1 M(say)
NaHCO₃ +  HCl  → No reaction
∴ Milliequivalents of Na₂CO₃ = Milliequivalents of HCl = x Using methyl orange,
NaHCO₃+  HCl → NaCl  + CO₂ +  H₂O
Milliequivalents of NaHCO₃ (from Na₂CO₃) + Milliequivalents of NaHCO₃ (original mixture) = y.
or  x + Milliequivalents of NaHCO₃ (original mixture) = y
Milliequivalents of NaHCO₃ (original mixture) = (y -x)
∴ Volume of HCl required = (y -x)ml.

Q.12. A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of a N/20 HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. Calculate the amount of KOH present in the solution.
(a) 0.014 g
(b) 0.24 g
(c) 0.038 g
(d) 5.4 g

Correct Answer is option (a)
With phenolphthalein, n_KOH + nNa₂CO₃ = 0.75 × 10^-3 ...(1)
With methyl orange, n_KOH + 2nNa₂CO₃  = 1.25 × 10^-3 ...(2)
From (1) and (2) n_KOH = 0.25 × 10^-3
Amount of KOH = 0.25 × 10^-3 × 56 = 0.014 g.

Q.13. 8.7 gm of pure MnO₂ is heated with an excess of HCl and the gas evolved is passed into a solution of KI. The amount of I₂ liberated is
(a) 0.6 mole
(b) 45.4 gm
(c) 25.4 gm
(d) 9.7 gm

Correct Answer is option (c)
MnO₂ + 4HCl → MnCl₂ + Cl₂ + H₂O
Cl₂ + 2KI → I₂ + 2 KCl
Moles of MnO₂ = 8.7/8.7 = 0.1
Moles of Cl₂ = 0.1
Moles of I₂ = 0.1
Mass of I₂ = 0.1 × 254 = 25.4 gm.

Q.14. Equivalent weight of HNO₃ in following reaction,
Cu  +  8HNO₃ → 3Cu(NO₃)₂ +  2NO  +  4H₂O is
(a) 63/3
(b) 73/5
(c) 4 × 63/3
(d) 43

Correct Answer is option (a)
From the given balanced equation, this is clear that only two moles of NO^-₃ undergo change in oxidation state while six moles remain in same oxidation state.
2HNO₃ +  6H⁺ +  6e^- → 2NO  +  4H₂O
Total 8 moles of HNO₃ exchange 6 moles of electrons.
1 mole of HNO₃ exchange 6/8 or 3/4 mole of electron
∴ n-factor of HNO₃ = 3/4.
Equivalent weight of HNO₃ = 63/3

Q.15. 3 mol of a mixture of FeSO₄ and Fe₂(SO₄)₃ required 100 ml of 2 M KMnO₄ solution in acidic medium. Hence the mole fraction of FeSO₄ in the mixture is
(a) 2/3
(b) 1/3
(c) 2/7
(d) 4/5

Correct Answer is option (b)
Equivalents of FeSO4 = Equivalents of KMnO4
= 100 × 10^-3 × 2 × 5 = 1
∴ Moles of FeSO₄ = 1 (∵ nFeSO₄ = 1)
∴ Mole fraction of FeSO₄ in the mixture = 1/3.