Test: JEE Main 35 Year PYQs- Some Basic Concepts of Chemistry - JEE MCQ

Empirical formula = C₃H₄N

(C₃H₄N)_n = 108, (12 × 3 + 4 × 1 + 14)_n =108

(54)_n = 108 ⇒ n = = 2

∴ molecular formula = C₆H₈N₂

Among all the given options molarity is correct because the term molarity involve volume which increases on increasing temperature.

Fe (no. of moles) = = 10 moles

C (no. of moles)  in 60 g of C = 60/12 = 5 moles.

2BCl₃ + 3H₂ → 2B+ 6HCl

Now, since 10.8 gm boron requires hydrogen

= x 22.4 L at N.T.P

hence 21.6 gm boron requires hydrogen

x 21.6 = 67.2 L at N.T. P..

25 × N = 0.1 × 35 ; N = 0.14
Ba(OH)₂ is diacid base hence N = M × 2  or  M =  ⇒  M = 0.07 M

Moles of urea present in 100 ml of sol.= 

= 0.01M

[ ∵ M = Moles of solute present in 1L of solution]

N₁V₁ = N₂ V₂ ( Note : H₃PO₃ is dibasic
∴ M = 2N) 20 x 0.2 = 0.1 x V (Thus. 0.1 M = 0.2 N)
∴ V = 40 ml

H₂SO₄ is dibasic. 0.1 M H₂SO₄ = 0.2N H₂SO₄ [ ∵ M = 2xN ]
Meq of H₂SO₄ taken = = 100 x 0.2 = 20
Meq of H₂SO₄ neutralised by NaOH = 20 x 0.5 = 10
Meq of H₂SO₄ neutralised by NH₃  = 20 – 10 = 10

% of  N₂ = 

= 46.6

% of nitrogen in urea= 46.6

[Mol.wt of urea =60]

Similarly % of Nitrogen in Benzamide

= 11.5%   [C₆H₅CONH₂ = 121]

Acetamide  = = 23.4%  [ CH₃CONH₂ = 59]

Thiourea =  = 36.8%  [NH₂CSNH₂ = 76]

Hence the compound must be urea.

TIPS/Formulae : From the molarity equation.
M₁V₁ + M₂V₂ = MV
Let M be the molarity of final mixture,

where V = V₁ + V₂

 = 1.344 M

Relative atomic mass

Now if we use 1/6 in place of 1/12 the formula becomes

Relative atomic m ass =

∴ Relative atomic mass decrease twice

1 Mole of Mg₃(PO₄)₂ contains 8 mole of oxygen atoms

∴  8 mole of oxygen atoms ≡ 1 mole of Mg₃(PO₄)₂ mole of Mg₃(PO₄)₂

0.25 mole of oxygen atom ≡ x mole of Mg₃(PO₄)₂
= 3.125x 10^-2  mole of  Mg₃(PO₄)₂

TIPS/Formulae : Apply the formula d = 

∴ 1.02 = 2.05 

On solving we get, m = 2.288 mol/kg

Since molarity  of solution is 3.60 M. It means 3.6 moles of H₂SO₄ is present in its 1 litre solution.
Mass of 3.6 moles of H₂SO₄ = Moles × Molecular mass = 3.6 × 98 g = 352.8 g
∴ 1000 ml solution has 352.8 g of H₂SO₄
Given that 29 g of H₂SO₄ is present in = 100 g of solution
∴ 352.8 g of H₂SO₄ is present in

 of solution = 1216 g of solution

Density =  = 1.216 g/ml  = 1.22 g/ml

2Al(s) + 6HCl(aq) → 2Al³⁺(aq) + 6Cl^–(aq) + 3H₂(g)
∵ 6 moles of HCl produces = 3 moles of H₂ = 3 × 22.4 L of H₂ at S.T.P
∴ 1 mole of HCl produces =  of H₂ at S.T..PP      
= 11.2 L of  H₂ at STP

On balancing the given equations we get

2MnO₄^- + 5C₂O₄^- + 16H + —→ 2Mn ++
+10CO₂+ 8H₂O
So, x = 2, y = 5 & z = 16

∵18 g, H₂O contains = 2 gm H

∴ 0.72 gm H₂O contains =  = 0.08 gm H

∵ 44 gm CO₂ contains = 12 gm C

∴ 3.08 gm CO₂ contains = = 0.84 gm C

∴ C : H =  = 0.07 : 0.08 = 7 : 8

∴ Empirical formula = C₇H₈

For a one mole of the oxide Moles of M = 0.98, Moles of O^2– = 1
Let moles of M³⁺ = x Moles of M²⁺ = 0.98– x on balancing charge (0.98 – x) × 2 + 3x – 2 = 0 ⇒ x = 0.04

 = 4.08%