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Test: Mole Concept, Volumetric & Redox - 1

20 Questions MCQ Test Physical Chemistry | Test: Mole Concept, Volumetric & Redox - 1

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Attempt Test: Mole Concept, Volumetric & Redox - 1 | 20 questions in 90 minutes | Mock test for Chemistry preparation | Free important questions MCQ to study Physical Chemistry for Chemistry Exam | Download free PDF with solutions

QUESTION: 1

4.4g of CO₂ and 2.24 litre of H₂at STP (273.15 K and 1 atm pressure) are mixed in a container. The total of molecules present in the container will be

A.
6.023 x 10²³
B.
12.044 x 10²²
C.
6.023 x 10²²
D.
6.023 x 10²⁴

Solution:

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

QUESTION: 2

4I^– + Hg²⁺ → HgI₄^2– ; 1 mole of each Hg²⁺ and I^– will form:

A.
1 mole of HgI₄^2–
B.
0.5 mole of HgI₄^2–
C.
0.25 mole of HgI₄^2–
D.
2 mole of HgI₄^2–

Solution:

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-

QUESTION: 3

How many grams of I₂ are present in a solution which requires 40 ml of 0.11N Na₂S₂O₃ to react with it via the reaction:
S₂O₃^2– + I₂ → S₄O₆^2– + 2I^–?

A.
12.7g
B.
0.558g
C.
25.4g
D.
11.4g

Solution:

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

QUESTION: 4

Two elements A (atomic mass = 75) and B (atomic mass = 16) combine to yield a compound. The % by weight of A in the compound was found to be 75.08. The formula of compound is:

A.
A₂B
B.
A₂B₃
C.
AB
D.
AB₂

Solution:

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.

QUESTION: 5

What weight of HNO₃ is needed to convert 5g of Iodine into Iodic acid according to the reaction:
I₂ + HNO₃→ HIO₃ + NO₂ +H₂O

A.
12.4g
B.
24.8g
C.
0.248g
D.
49.6g

Solution:

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

QUESTION: 6

In the reaction, VO + Fe₂O₃ → FeO + V₂O₅, the equilvalent weight of V₂O₅ is equal to its

A.
Mol. Wt.
B.
Mol. Wt./4
C.
Mol. Wt./6
D.
Mol. Wt./8

Solution:

QUESTION: 7

When BrO₃^– ion reacts with Br^– ion in acid solution Br₂ is liberated. The eq. weight of KBrO₃ in this reaction is:

A.
M/8
B.
M/3
C.
M/5
D.
M/6

Solution:

10e + 2Br⁵⁺→ Br⁰₂

QUESTION: 8

The hydrated salt, Na₂SO₄.nH₂O, undergoes 55.9 % loss in weight on heating and become anhydrous. The value of n will be:

A.
5
B.
3
C.
7
D.
10

Solution:

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

In 100g of the compound, 44.1 g Na₂SO₄contains 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.

QUESTION: 9

When a metal is burnt, its weight is increased by 24%. The eq. weight of the metal will be:

A.
25
B.
24
C.
33.3
D.
76

Solution:

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

QUESTION: 10

One mole of potassium chlorate is thermally decomposed and excess of aluminum is burnt in the gaseous product .How many moles of aluminum oxide are formed?

A.
1
B.
1.5
C.
2
D.
3

Solution:

2KClO₃→ 2KCl + 3O₂

4Al + 3O₂→ 2Al₂O₃

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

Thus, if 1 mole of KClO₃is decomposed, 1.5 moles of O₂is 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 Al₂O₃produced =

QUESTION: 11

How much Cl₂ at STP is liberated when one mole of KMnO₄ reacts with HCl ?

A.
11.2 L
B.
22.4 L
C.
44.8 L
D.
56 L

Solution:

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

∴ Volume of 2.5 moles of Cl₂= 56 litres

QUESTION: 12

Minimum quantity of H₂S needed to precipitate 63.5g of Cu²⁺ is nearly:

A.
63.5 g
B.
34g
C.
20g
D.
126g

Solution:

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

QUESTION: 13

2g of CaCO₃ was treated with 0.1M HCl (500 ml). The volume of CO₂ evolved at STP after heating the solution is:

A.
0.448 L
B.
0.224 L
C.
4.48 L
D.
44.8 L

Solution:

QUESTION: 14

10g of CaCO₃ on heating gives 5g of the residue (as CaO). The % yield of the reaction is approx:

A.
50%
B.
72%
C.
89 %
D.
100%

Solution:

QUESTION: 15

‘x’ g of KClO₃ on decomposition gives ‘y’ ml of O₂ at STP. The % purity of KClO₃ would be

Solution:

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)

QUESTION: 16

33.6g of an impure sample of sodium bicarbonate when heated strongly gave 4.4g of CO₂. The % purity of NaHCO₃ would be:

A.
25%
B.
50%
C.
75%
D.
100%

Solution:

Balanced Equation:
2NaHCO₃= Na₂CO₃ + CO₂ + H₂O

Moles of CO₂ = 4.4/44 =0.1
∵ 1 mole of CO₂is 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%.

QUESTION: 17

0.16g of dibasic acid required 25ml of N/10 NaOH for complete neutralization. Molecular wt. of acid is:

A.
32
B.
64
C.
128
D.
256

Solution:

Dibasic acid NaOH;

N₁V₁= N₂V₂

M = 2 × E = 2 × 64 = 128

QUESTION: 18

The molar coefficients of AsO₃^3– and MnO₄^- in the reaction are:

AsO₃^3– + MnO₄^Θ → AsO₄^3– + MnO₂

A.
2, 3
B.
3, 2
C.
2, 2
D.
3, 3

Solution:

QUESTION: 19

Which of the following is a disproportionation reaction?

A.
CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂
B.
2CrO₄^2– + 2H⁺ → Cr₂O₇^2– + H₂O
C.
Cr₂O₇^2– + 2OH^– → 2CrO₄^2– +H₂O
D.
Cu₂O + 2H⁺ → Cu + Cu²⁺ + H₂O

Solution:

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)

QUESTION: 20

Hydrogen and oxygen combine to form H₂O₂ and H₂O containing 5.93% and 11.2 % hydrogen respectively. The data illustrates:

A.
Law of conservation of mass
B.
Law of constant proportion
C.
Law of reciprocal proportion
D.
Law of multiple proportion

Solution:

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.