Chemistry Test 129

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Chemistry Test 129

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Weekly Quiz Competition

Question 1
4 / -1

Two oxides of metal contain 27.6℅ and 30℅ oxygen respectively. If formula of first oxide is M₃O₄, then formula of second oxide is:

A
MO₂

B
M₂O₃

C
M₂O₅

D
MO

Solution

Given that, formula of first oxide = M₃O₄

Let mass of the metal = x

% of metal in M₃O₄ = (3x / 3x + 64) * 100

but as given % age = (100 - 27.6) = 72.4%

so, (3x / 3x + 64) * 100 = 72.4

or x = 56.

in 2nd oxide,

oxygen = 30%....so metal = 70%

so, the ratio is

M : O

70 / 56 : 30 / 16

1.25 : 1.875

2 : 3

so, 2nd oxide is M₂O₃
Question 2
4 / -1

5.6 litres of a gas at N.T.P. are found to have a mass of 11 g. The molecular mass of the gas is

A
22

B
44

C
88

D
32

Solution

5.6lit = 11g

22.4lit = ? (x) g

x = 22.4 × 11 / 5.6

x = 44g

Thus, Molecular mass of gas is 44g
Question 3
4 / -1

The crystalline salt, Na₂SO₄xH₂O on heating loses 55.9% of its weight. The formula of the crystalline salt is

A
Na₂SO₄.5H₂O

B
Na₂SO₄.7H₂O

C
Na₂SO₄.10H₂O

D
None of these

Solution

Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18

= 142 + 18x

On heating it will lose water and become anhydrous. 55.9% mass is the mass of water in Na₂SO₄.xH2O.

Mass by mass% = (mass / molar mass of compound) × 100

mass of water = 18x

55.9 = (18x / 142 + 18x) × 100

55.9(142 + 18x) = 1800x

7937.8 + 1006.2x = 1800x

x = 9.99

x = 10 (approx.)

So, the molecular formula of compound is Na₂SO₄.10H₂O.
Question 4
4 / -1

0.5 g of a metal on oxidation gave 0.79 g of its oxide. The equivalent weight of the metal is (approximately)

A
10

B
14

C
20

D
40

Solution

Given,

Mass of metal = 0.5 g

Mass of oxide = 0.79 g

First we have to calculate the Mass of Oxygen in metal oxide.

Mass of Oxygen = Mass of oxide - Mass of Metal = 0.79 - 0.5 = 0.29 g

The equivalent weight of oxygen is 8 g/eq.

0.29 g of oxygen combines with the 0.5 g of metal

8 g of oxygen combines with of metal.

Therefore, the equivalent weight of metal is 13.793 g/eq.
Question 5
4 / -1

Air contains nearly 20% oxygen by volume. The volume of air needed for complete combustion of 100 cc of acetylene will be

A
500 cc

B
1250 cc

C
250 cc

D
100 cc

Solution

Equation

2 C₂H₂ + 5O₂ → 4CO₂ + 2 H₂O

In this reaction, we can see that, 2 moles of C₂H₂reacts with 5 moles of O₂

Or, 1 mole of C₂H₂ reacts with 5 / 2 moles of O₂

Under standard conditions, we can say that:

22.44 L of C₂H₂ reacts with (5 / 2) x 22.4 L moles of O₂

So, 100 cc of C₂H₂ will reacts with (5 / 2) x 100 cc = 250 cc of O₂.

But, since air contains 20 % of oxygen by volume, the amount of air needed to react with 100 cc of C₂H₂ will be = 250 x (100 / 20) = 1250 cc of air.
Question 6
4 / -1

A solution of 10 ml(M/10) FeSO₄ was titrated with KMnO₄ solution in acidic medium. The amount of KMnO₄ used will be

A
5 ml of 0.1 M

B
10 ml of 0.1 M

C
10 ml of 0.5 M

D
10 ml of 0.2 M

Solution

The equation of the Mn²⁺ in acidic medium will be.

MnO₄^- + 8H⁺+ 5 Fe²⁺ → Mn²⁺+ 5Fe³⁺ +4H₂O.

We can balance the reaction by using redox balancing of acid and base.

So, we know from the equation that 5 moles of iron react to form 1 mole of Mn⁺².

Thus, we can say that (10*0.1) moles of iron will react with the (10*0.1/5) moles of Manganese ions, which will be 0.2 thus, 10ml volume of KMnO4 and 0.02M moles of KMnO4. We will get the same result if we use M₁V₁ = M₂V₂ formulae.
Question 7
4 / -1

At S.T.P. the density of CCl₄ vapour in g/L will be nearest to

A
6.84

B
3.24

C
10.26

D
4.57

Solution

Molecular Weight of CCl₄ = 35.5 + 12 x 4 = 154

1 mole CCl₄ vapour=154 g

According to Avogadro’s hypothesis we know that the molar volume of a gas at STP is 22.4 litres

= 154 / 22.4 = 6.87
Question 8
4 / -1

The law of multiple proportions was proposed by

A
Lavoisier

B
Dalton

C
Proust

D
Gay-Lussac

Solution

Dalton's law of multiple proportions is part of the basis for modern atomic theory, along with Joseph Proust's law of definite composition (which states that compounds are formed by defined mass ratios of reacting elements) and the law of conservation of mass that was proposed by Antoine Lavoisier.
Question 9
4 / -1

The anode rays were discovered by

A
Goldstein

B
Crooks

C
Chadwick

D
Thomson

Solution

The discovery of protons can be attributed to Rutherford. In 1886 Goldstein discovered existence of positively charged rays in the discharge tube by using perforated cathode. These rays were named as anode rays or cannal rays.
Question 10
4 / -1

Which of the following transitions of electrons in the hydrogen, atom will emit maximum energy ?

A
n₅ → n₄

B
n₄ → n₃

C
n₃ → n₂

D
All will emit same energy

Solution

Energy is given by formula

E = hc / λ

1 / λ = RZ² [ 1 / n²₁ - 1 / n₂² ]

E = RZ² hc [ 1 / n²₁ - 1 / n²₂]

This energy will be highest when the transition is from n₃ → n₂.

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Chemistry Test 129

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Question 1

MO2

M2O3

M2O5

MO

Solution

Question 2

22

44

88

32

Solution

Question 3

Na2SO4.5H2O

Na2SO4.7H2O

Na2SO4.10H2O

None of these

Solution

Question 4

10

14

20

40

Solution

Question 5

500 cc

1250 cc

250 cc

100 cc

Solution

Question 6

5 ml of 0.1 M

10 ml of 0.1 M

10 ml of 0.5 M

10 ml of 0.2 M

Solution

Question 7

6.84

3.24

10.26

4.57

Solution

Question 8

Lavoisier

Dalton

Proust

Gay-Lussac

Solution

Question 9

Goldstein

Crooks

Chadwick

Thomson

Solution

Question 10

n5 → n4

n4 → n3

n3 → n2

All will emit same energy

Solution

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MO₂

M₂O₃

M₂O₅

Na₂SO₄.5H₂O

Na₂SO₄.7H₂O

Na₂SO₄.10H₂O

n₅ → n₄

n₄ → n₃

n₃ → n₂