Atoms and Molecules Test

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Atoms and Molecules Test - 63

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Weekly Quiz Competition

Question 1
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Who gave the definition of an element?

A
Robert Boyle

B
John Dalton

C
Lavoisier

D
Thomson

Solution

The word element was first given by Rebert Boyle. It is defined as the simplest chemical substance that cannot be broken down during a chemical reaction.

So, the correct option is $$A$$
Question 2
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$$12$$ gm of an alkaline earth metal gives $$14.8$$g of its nitride. The atomic mass of metal is:

A
$$12$$

B
$$24$$

C
$$20$$

D
$$40$$

Solution

Let the alkaline earth metal be $$A$$.

$$\therefore 3A+N_{2}\rightarrow A_{3}N_{2}$$

$$\Rightarrow \dfrac{Molecular\ mass\ of\ compound}{Atomic\ mass\ of\ metal}$$

$$=\dfrac{Wt.\ of\ compound}{Wt.\ of\ metal}$$

Let the atomic mass of metal be $$x$$.

$$\rightarrow$$ Molecular mass of compound$$=3x+28$$

$$\Rightarrow \dfrac{3x+28}{3x}=\dfrac{14.8}{12}$$

$$\Rightarrow x=40$$
Question 3
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A compound was found to contain $$5.37\%$$ nitrogen by mass. What is the minimum molecular weight of compound?

A
$$26.07$$

B
$$2.607$$

C
$$260.7$$

D
None

Solution
Question 4
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A compound possesses 8% sulphur by mass. The least molecular mass is

A
100

B
600

C
400

D
200

Solution

Hint: The minimum number of Sulphur atoms which a molecule can possess is one.
Step 1: Finding the molecular mass of the compound.

$$Mass\ of\ 1\ sulphur\ atom\ =\ 32 \ grams.$$
$$ As\ the\ compound\ has\ 8\% \ sulphur\ by\ mass,$$
$$\therefore 8 \ grams\ of\ sulphur\ is\ present\ in\ 100 \ grams\ of\ compound.$$
$$\therefore 32 \ grams\ of\ sulphur\ is\ present\ in\ \dfrac{32}{8} \times 100=400\ grams \ of\ compound.$$

$$Hence,\ the\ least\ molecular\ mass\ of\ the\ compound\ is\ 400 \ grams.$$
Final Step: Correct answer - (C) The least molecular mass of the compound is $$400 \ grams.$$
Question 5
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48 g of Mg contain the same number of atoms as 160 g another mono-atomic element.The atomic mass the element is:

A
24

B
320

C
80

D
40

Solution
Question 6
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A sample of pure carbon dioxide, irrespective of its source contains $$27.27 \%$$ carbon and $$72.73 \%$$ oxygen. The given data supports:

A
Law of constant proportions

B
Law of conservation of mass

C
Law of reciprocal proporties

D
Law of multiple proportions

Solution

The ratio by mass in which carbon and oxygen present in $$CO_2$$:
$$\dfrac{mass \ of \ carbon} {mass \ of \ oxygen} $$=$$ \dfrac{12}{2×16}$$

Therefore the ratio by mass in $$CO_2$$ becomes $$3:8$$

Now we are given the percentage of carbon and oxygen in $$CO_2$$:
Now let us find the ratio by percentage in which carbon and oxygen are present in $$CO_2$$.
$$\dfrac{carbon \%} {oxygen \%} = \dfrac{27.27} {72.73}$$

As both the value is divisible by we get:
$$carbon: oxygen =3:8$$

Carbon dioxide prepared by any method contains carbon and oxygen in the same proportion by mass. Thus, we can say that the given data proves the law of constant proportions. The Law of constant proportion is also known as the law of definite proportion.
Hence, option A is correct.
Question 7
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In $$\overset { 14 }{ \underset { 7 }{ N } } $$ if mass attributed to electron were doubled & the mass attributed to protons were halved, the atomic mass would become approximately:-

A
Halved

B
Doubled

C
Reduced by 25%

D
Remain same

Solution
Question 8
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If Avagadro Number $${N}_{A}$$ is changed from $$6.022\times{10}^{23}{mol}^{-1}$$ to $$6.022\times{10}^{20}{mol}^{-1}$$ this would change

A
The ratio of elements to each other in a compound

B
The definitiion of mass in units of grams

C
The mass of one mole of carbon

D
The ratio of chemical species to each other in a balanced equation

Solution

Avogadro number:
If Avogadro number $$N_A$$ is changed from $$6.022 \times 10^{23}mol^{-1}$$ to $$6.022 \times 10^{23}mol^{-1}$$. This would change the mass of one mole of carbon 12g $$\rightarrow$$ 12mg i.e 1 mol of carbon atom mass = 0.012gm.
Hence option $$C$$ is correct.
Question 9
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The weight of a molecule of compound $${C_{60}}{H_{22}}$$ is

A
$$1.231 \times {10^{ - 21}}g$$

B
$$24 \times {10^{ - 21}}g$$

C
$$5.025 \times {10^{23}}g$$

D
$$16.023 \times {10^{23}}g$$

Solution
Question 10
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Iron forms two oxides, in first oxide $$56$$ grams of Iron is found to be combined with $$16$$ gram oxygen and in second oxide $$112$$ gram of Iron is found to be combined with $$48$$ gram oxygen. This data satisfy the law of

A
Conservation of mass

B
Reciprocal proportion

C
Multiple proportions

D
Combining volume

Solution