JEE Advanced (One or More Correct Option): Some Basic Concepts of Chemistry

Q.1. In which of the following statements is/are correct?
(a) Weight of 12.044 × 10²³ atoms of carbon is 24g.
(b) Weight of 6.022 × 10²³ molecules of CaCO₃ is 100 g.
(c) Number of moles of 0.635 g  of Cu is 2. (At wt of Cu = 63.5)
(d) The number of molecules in 11.2 litre of SO2 gas at NTP is 0.5.

Correct Answer is option (a, b and d)
For (A) → The number of moles of

Wt of  C-atm = 2 × 12 = 24g
For (B) weight of CaCO₃ = number of moles × mol.wt

For (C) number of moles of Cu

Hence it is wrong.
For (D) number of

Q.2. 4g of hydrogen is ignited with 4 g oxygen. Following reaction takes place
2H₂ + O₂ → 2H₂O
Select the correct  statement(s).
(a) Oxygen is limiting reactant
(b) Hydrogen is limiting reactant
(c) 4g of hydrogen reacts with 2g oxygen
(d) 4.5 g of water will be formed

Correct Answer is option (a and d)

4 g of H₂ will completely react with 32 g of O₂ , but only 4 g of O₂ is present.
Hence, O₂ is limiting reactant.
Since, 32 g of O₂ produces 36g of H₂O
Hence 4 g of O₂ will produce 36/32 × 4 = 4.5g of H₂O

Q.3. 0.2 mole of Na₃PO₄ and 0.5 mole of Ba (NO₃)₂ are mixed in one litre of solution. Which of the following is/are correct.
(a) 0.2 mole of  Ba₃ (PO₄)₂ is obtained
(b) 0.1 mole of  Ba₃ (PO₄)₂ is obtained
(c) Molarity  of Ba₃ (PO₄)₂ in solution  is 0.1 M
(d) Molarity  of Ba₃ (PO₄)₂ in solution  is 0.2 M

Correct Answer is option (b and c)
2Na₃PO₄ + 3Ba (NO₃)₂ → Ba₃ (PO₄)₂ + 6NaNO₃

Na₃PO₄ is limiting reactant.
∴ Moles of Ba₃(PO₄)₂ formed = 0.1
Molarity of Ba₃ (PO₄)₂ = 0.1/1 = 0.1M

Q.4. 1 mole of H₂SO₄ will exactly neutralise
(a) 2 moles of NH₄OH
(b) 1 mole of Ca (OH)₂
(c) 2 moles of NaOH
(d) 0.5 mole of Ba (OH)₂

Correct Answer is option (a, b and c)


Equal number of equivalents of acid and base neutralises.

Q.5. 200 ml of 0.3 M Ca(OH)₂ will be completely neutralized by
(a) 1200 ml of 0.1 M HCl
(b) 600 ml of 0.1 M H₂SO₄
(c) 400 ml of 0.1 M H₃PO₄
(d) 600 ml of 0.2 M HNO₃

Correct Answer is option (a, b, c and d)
m eq of Ca(OH)₂ ⇒ 200 × 0.3 × 2 ⇒ 120 (n_f = n-factor)
(A) 1200 × 0.1 = 120 M eq of HCl HCl n_f = 1
(B) 600 × 0.1 × 2 = 120 M eq of H₂SO₄ H₂SO₄ n_f = 2
(C) 400 × 0.1 × 3 = 120 M eq of H₃PO₄ H₃PO₄ n_f = 3
(D) 600 × 0.2 × 1 = 120 M eq of HNO₃ HNO₃ n_f = 1

Q.6. 100 ml of 0.1 M H₂SO₄ is mixed with 200 ml of 0.2 M HCl then resulting mixture will be neutralized by
(a) 600 ml of 0.1 M NaOH
(b) 300 ml of 0.1 M Ca(OH)₂ 
(c) 200 ml 0.1 M of Al(OH)₃ 
(d) 400 ml of 0.1 M KOH

Correct Answer is option (a, b and c)
m eq of H₂SO₄ + m eq of HCl
⇒ 100 × 0.1 × 2 + 200 × 0.2 × 1
⇒ 20 + 40  = 60  m eq of Acid
(A) 600 × 0.1 = 60 m eq of NaOH
(B) 300 × 0.1 × 2 = 60 m eq of Ca(OH)₂
(C) 200 × 0.1 × 3 = 60 m eq of Al(OH)₃
(D) 400 × 0.1 × 1 = 40 m eq of KOH

Q.7. 1 molar of 1 litre of H₂SO₄ will exactly neutralize
(a) 2 molar of 1 litre NH₃
(b) 1 molar of 1 litre Ca (OH)₂
(c) 0.5 molar of 1 litre Ba (OH)₂
(d) 2 molar of 1 litre NaOH

Correct Answer is option (a, b and d)
n-factor for H₂SO₄ = 2
Equivalent of H₂SO₄ = 2
Equivalence of NH₃ = 2 × n - factor = 2 × 1 = 2
Equivalence of Ca(OH)₂ = 1 × 2 = 2
Equivalence of Ba(OH)₂ = 0.5 × 2 = 1
Equivalence of NaOH =2 × 1 = 2

Q.8. 10 moles of SO₂ and 4 moles of O₂ are mixed in a closed vessel. Following reaction takes place 2SO₂ (g) + O₂ (g) → 2SO₃ (g). Select the correct statement(s).
(a) SO₂ is limiting reactant
(b) O₂ is limiting reactant
(c) 8 moles of SO₃ are formed
(d) 10 moles of SO₃ are formed

Correct Answer is option (b and c)
2SO + O₂ → 2SO₃
Moles taken initially 10 moles 4 moles 0
Moles involved in reaction 8    4    8
Final moles 2    0     8
O is limiting reactant. SO₂ is excess reactant as 2 moles of SO₂ remains unreacted. 8 moles of SO₃ are produced.

Q.9. Which of following contains the same number of moles?
(a) 1g of O₂ and 2 g of SO₂ 
(b) 1 g of O₂ and 1 g of O₃ 
(c) 1 g of CO₂ and 1g of N₂O
(d) 11.2 L of CO₂ at STP and 1 g H₂

Correct Answer is option (a, c and d)
Moles in 1 g O₂ = 1/32
Moles  in 2 g SO₂ = 2/64 = 1/32
Moles  in 1 g O₃ =1/48
Moles in 1 g CO₂ = 1/44
Moles  in 1g N₂O = 1/44
Moles  in  11.2 L CO₂ (STP) = 11.2/22.4 = 1/2
Moles  in 1g H₂ = 1/2

Q.10. A sample of H₂O₂ solution labelled as "28 volume" has density of 26.5 g/L. Mark the correct option(s) representing concentration of same solution in other units:
(a) H₂O₂ M = 2.5
(b) %w/v = 17
(c) Mole fraction of H₂O₂ = 0.2
(d) m H₂O₂ = 13.88

Correct Answer is option (a, c and d)
V_strength = 56

∴ 1 L contain 2.5 moles of H₂O₂
Or 2.5 × 34 = 85g H₂O₂
Wt. of 1 litre solution = 265 g (∵ d = 265g/L)
∴ wH₂O  = 180g or moles of H₂O = 10