Test: Mole Concept, Volumetric & Redox - 1 - Chemistry MCQ

Number of moles of CO₂ =

∵ 22.4 L of H₂ has 1 mole of H₂

∴ The number of moles in 2.24L 
The total moles are 0.2.
⇒ Total molecules in 0.2 moles 

Balanced reaction is as follows :
Hg²⁺(aq) + 4I^- ​ (aq) ⇋​   HgI₄^2-​ (aq)​

∵ 4 mole of I^- produces = 1 mole of HgI₄^2-​​
∴ 1 mole I^- produces = 1/4 moles of HgI₄^2-​​

Meq. of I₂ = Meq. of Na₂S₂O₃ = 40 × 0.11
∴ (W × 2 × 1000)/254  = 40 × 0.11
W of I₂ = 0.558g

Molecular weight of A₂B₃ = 75 × 2 + 16 × 3 = 198
198g of A₂B₃ =150g of A
∴ 10g of A₂B₃ =   = 75.08g of A

So option B is correct. 

The balanced reaction is:
I₂ + 10HNO₃ → 2HIO₃ + 10NO₂ + 4H₂O

(127 * 2)g of iodine requires (63 * 10)g of HNO₃
Thus, 5g of I₂ requires
= [(63 * 10)/(127 * 2)] * 5 = 12.4g 

10e + 2Br⁵⁺ → Br⁰₂

Molecular Mass of Na₂SO₄ is = 2 x 23 + 32 + 4 x 16 = 142 gm.

In 100g of the compound, 44.1 g Na₂SO₄ contains 55.9 g of H₂O.
(∵ 55.9 % is water)
⇒ 1g Na₂SO₄ contains = (55.9/44.1)g H₂O
∴ 142 g Na₂SO₄ contains = (55.9/44.1) x 142 = 180 g H₂O .

Hence, the number of water molecules = 180/18 = 10
(Since the Molecular mass of H₂O is 18).

Thus, the value of n is 10 and the formula is Na₂SO₄.10H₂O.

Equivalent weight of O^2-(oxide) ion is: Molar mass of oxygen/charge on oxygen
= 16/2 = 8 g eq^-1

One equivalent of the metal reacts with one equivalent of oxygen to form metal oxide.

Let M be the equivalent weight of the metal.
Therefore, M grams of the metal will react with 8g of oxygen.
The corresponding metal oxide will have a mass of 8 grams more than the mass of metal M. This 8 gram is 24% of M. 
i.e. M x 24/100 = 8 gram
Thus, M = 100 x 8/24 = 33.33 g

2KClO₃ → 2KCl + 3O₂

4Al + 3O₂ → 2Al₂O₃

The molar ratio for the first reaction is 2:3 for KClO₃:O₂

Thus, if 1 mole of  KClO₃ is decomposed, 1.5 moles of O₂ is produced.

The molar ratio for the second reaction is: 3:2 for O₂:Al₂O₃. 

3 moles of oxygen produces 2 moles of aluminium oxide. 

Since, we have 1.5 moles of oxygen, number of moles of Al₂O₃  produced =

Potassium permanganate reacts with hydrochloric acid according to the following equation:
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂

⇒ 2 moles of KMnO₄ produces 5 moles of Cl₂
⇒ 1 mole of KMnO₄ will produce 2.5 moles of Cl₂

1 mole of CO₂ will have 22.4 litre sof volume
∴ Volume of 2.5 moles of Cl₂ = 56 litres

Chemical Reaction: Cu²⁺+ H₂S → CuS + 2H⁺

In this reaction, to precipitate 1 mol of Cu²⁺ we require 1 mol of H₂S.
Moles of Cu²⁺ = 63.5 g / 63.5 g mol^-1 = 1 mol
Hence, to precipitate 1 mol of Cu²⁺ we require 1 mol of H₂S
i.e. Mass of H₂S = 1 mol x 34 g mol^-1 = 34 g

The balanced eqn. will be:
2KClO₃ = 2KCl+3O₂

The no. of moles of Oxygen = y/22.4
3 moles of Oxygen is produced from 2 moles of KClO₃
⇒ 1 mole will be produced from 2/3 moles of KClO₃
∴ y/22.4 will be producing from (2/3) × (y/22.4)
Hence, the mass of KClO₃ used = (2/3) × (y/22.4)×M

% Purity = (2 × y × M)/(3 × 22.4 × x)

Balanced Equation:
2NaHCO₃ = Na₂CO₃ + CO₂ + H₂O

Moles of CO₂ = 4.4/44 =0.1
∵ 1 mole of CO₂ is produce by 2 Moles of NaHCO₃
⇒ Moles of NaHCO₃ =0.1×2/1 =0.2
∴ Wt. of NaHCO₃ = 0.2 × 84 = 16.8gm
% Purity of NaHCO₃ = (16.8/33.6) × 100 = 50%.

Dibasic acid NaOH; 

N₁V₁ = N₂V₂

M = 2 × E = 2 × 64 = 128

.

A disproportionation reaction is a reaction in which a single element undergoes reduction and oxidation

i.e in option d), Cu has changed its oxidation state from +1 to 0 (i.e reduction) and +1 to +2 (i.e oxidation)

For predicting the law of chemical combination, we have to have find out the ratio of masses in both cases:
Let the total mass in each case be 100 grams.

For H₂O₂
​​Mass of hydrogen = 5.93 gm
Mass of oxygen = 100-5.93 = 94.07 gm
The ratio of the mass of hydrogen to the mass of oxygen, H:O = 5.93 : 94.07 = 1 : 16

For H₂O
​Mass of hydrogen = 11.2 gm
Mass of oxygen = 100-11.2 = 88.8 gm
The ratio of the mass of hydrogen to the mass of oxygen, H:O = 11.2 : 88.8 = 1 : 32

As the ratio of the mass of hydrogen remains the same in both cases and the ratio of masses of oxygen in both cases (16: 32 or 1: 2) is a whole number ratio.

Therefore, the above data illustrate the law of multiple proportions.