Atoms and Molecules Test

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Atoms and Molecules Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

A sample of $$CaCO_3$$ has Ca - 40%, C = 12% and O = 48%. If the law of constant proportions is true, then the mass of Ca in 5 g of $$CaCO_3$$ from another source will be:

A
2.0 g

B
0.2 g

C
0.02 g

D
20.0 g

Solution

Since, mass percentage of Ca in $$CaCO_3 = 40 \text{%} $$ and law of constant proportion is true.
So, the mass percentage of $$Ca $$ in $$ 5gm$$ $$CaCO_3$$ will also be $$40\text%.$$

Hence,$$ 40\text{% of 5gm =}$$$$ 5\times $$$$\dfrac{40}{100}$$ $$= 2gm $$

Hence, answer is option A.
Question 2
1 / -0

The number of electrons in one molecule of $$CO_{2}$$ are:

A
$$22$$

B
$$44$$

C
$$66$$

D
$$88$$

Solution

One molecule of $$CO_2$$ have $$22$$ electrons.

$$C + O+ O\Rightarrow 6 + 8+ 8 =22$$
Question 3
1 / -0

If one mole contains $$1.0 \times 10^{24}$$ particles, the mass of one mole of oxygen is:

A
53.2 g

B
5.32 g

C
32.0 g

D
16.0 g

Solution

1 mole of oxygen contains $$ \displaystyle 6.02 \times 10^{23}$$ particles and its mass is 32 g.
If a mole was to contain $$1.0 \times 10^{24}$$ particles, the mass of one mole of oxygen is
$$ \displaystyle \dfrac {32}{6.02 \times 10^{23}} \times 1.0 \times 10^{24} = 53.2$$ g.
Question 4
1 / -0

The law of constant proportions was enunciated by:

A
Dalton

B
Berthelot

C
Avogadro

D
Proust

Solution

Proust defined the law of constant proportions. It is also called as $$\text{Proust's law}$$. it states that "In a chemical substance the elements are always present in a definite proportion by mass".
Hence, the answer is option $$D$$.
Question 5
1 / -0

Zinc sulphate contains 22.65% Zn and 43.9% $$H_2O$$. If the law of constant proportions is true, then the mass of zinc required to give 40 g crystals will be:

A
90.6 g

B
9.06 g

C
0.906 g

D
906 g

Solution

$$ZnSO_4-22.65\% Zn $$ $$43.9\% H_2O$$

$$100g$$ crystal has $$22.65$$ $$Zn$$

$$40g$$ crystal has $$x$$ $$Zn$$

$$x=\cfrac{40\times 22.65}{100}=\cfrac{906}{100}=9.06g$$
Hence, mass of $$Zinc$$ required is $$9.06g$$.
Question 6
1 / -0

Law of constant composition is same as the law of:

A
conservation of mass

B
conversation of energy

C
multiple proportion

D
definite proportion

Solution

The law of constant composition is not the same as the law of conversation of mass or the law of conservation of energy, as these are related to the mass and energy of a system. It is also not related to the law of multiple proportions, which applies when two or more elements/ compounds have multiple ways of combining into different compounds.
The Law of Constant Composition, discovered by Joseph Proust, is also known as the Law of Definite Proportions.
According to the law of constant composition - "A chemical compound always contains its component element in the fixed ratio by mass." Hence option D is correct.
Question 7
1 / -0

Atomic size is of the order of____________.

A
$${10}^{-7} cm$$

B
$${10}^{-10} cm$$

C
$${10}^{-13} cm$$

D
$${10}^{-6} cm$$

Solution

The atomic size or atomic radius is of the order of $$\displaystyle {10}^{-7} cm$$ or $$\displaystyle {10}^{-9} m$$ or 1 nanometer(nm).
Question 8
1 / -0

Hydrogen combines with chlorine to form HCI. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as:

A
23 : 35.5

B
35.5 :23

C
1 : 1

D
23 : 1

Solution

Since, mass ratio of $$ H : Cl = 1 : 35.5$$ in HCl and $$Na : H = 23 : 1$$ in NaH.
And since according to law of constant proportion compounds always contains its elements in a fixed ratio by mass.
So, for$$\text{ 1gm of H}$$ we are having$$\text{ 35.5gm Cl and 23gm Na.}$$
Hence, the ratio $$Na : Cl = 23 : 35.5$$.
Hence, the answer is option A.
Question 9
1 / -0

The haemoglobin from red corpuscles of most mammals contains approximately 0.33% of iron by weight. The molecular weight of haemoglobin is 67,200. The number of iron atoms in each molecule of haemoglobin is:

A
1

B
2

C
3

D
4

Solution

$$100g$$ sample of hemoglobin contains 0.33 g iron
67200g sample of hemoglobin contains $$= 0.33\times67200/100 = 221.8g$$
$$\therefore$$ number of iron atoms per molecule of hemoglobin $$= 221.8/56 = 4 $$
Question 10
1 / -0

$$X\ gm$$ $$A$$ atoms on combining with $$Y$$ atoms of $$B$$ form 5 molecules of a compound containing $$A$$ & $$B$$. Find the molecular weight of compound formed. (Atomic weight of $$B = M$$)

A
$$\dfrac { \left( X{ N }_{ A }+MY \right) }{ 5 } $$

B
$$\dfrac { X+M }{ 5 } \\ $$

C
$$\dfrac { X+MY }{ 5 } $$

D
$$\left( \dfrac { X+MY{ N }_{ A } }{ 5 } \right) $$

Solution

Molecular weight of compound formed be $$W$$
$$X+MY =5W$$
$$W= \dfrac{X+MY}{5}$$

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Re-Attempt Weekly Quiz Competition

Atoms and Molecules Test - 56

Result

Atoms and Molecules Test - 56

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SHARING IS CARING

Question 1

2.0 g

0.2 g

0.02 g

20.0 g

Solution

Question 2

$$22$$

$$44$$

$$66$$

$$88$$

Solution

Question 3

53.2 g

5.32 g

32.0 g

16.0 g

Solution

Question 4

Dalton

Berthelot

Avogadro

Proust

Solution

Question 5

90.6 g

9.06 g

0.906 g

906 g

Solution

Question 6

conservation of mass

conversation of energy

multiple proportion

definite proportion

Solution

Question 7

$${10}^{-7} cm$$

$${10}^{-10} cm$$

$${10}^{-13} cm$$

$${10}^{-6} cm$$

Solution

Question 8

23 : 35.5

35.5 :23

1 : 1

23 : 1

Solution

Question 9

1

2

3

4

Solution

Question 10

$$\dfrac { \left( X{ N }_{ A }+MY \right) }{ 5 } $$

$$\dfrac { X+M }{ 5 } \\ $$

$$\dfrac { X+MY }{ 5 } $$

$$\left( \dfrac { X+MY{ N }_{ A } }{ 5 } \right) $$

Solution

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