Atoms and Molecules Test

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Atoms and Molecules Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

A gaseous mixture contains oxygen and nitrogen in the ratio of 1: 8 by mass. Therefore, the ratio of their respective number of molecules is:

A
1 : 8

B
1 : 1

C
7 : 64

D
1 :2

Solution

No of Molecules = $$\dfrac{Weight}{Gram\space Molecular\space Weight} \times {N}_{A}$$

Molecular wt of Hydrogen $${N}_{2}$$ = 28
Molecular wt of Oxygen $${O}_{2}$$ = 32

So Ratio of Number of molecules = $$\dfrac{1}{32} : \dfrac{8}{28}$$ = $$7 : 64$$
Question 2
1 / -0

If we consider that $$\frac{1}{6}$$, in place of $$\frac{1}{12}$$; mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

A
Decrease twice

B
Increase two fold

C
Remain unchanged

D
Be a function of the molecular mass of the substance

Solution
Question 3
1 / -0

What is the molecular mass of sodium sulphate?

A
154

B
119

C
165

D
142

Solution
Question 4
1 / -0

The molecular mass of $$K_2CO_3$$ is:

A
128 u

B
112 u

C
116 u

D
138 u

E
90 u

Solution

$$Molecular\ mass\ of\ K_2CO_3\ = (39 \times 2) + 12 + (16 \times 3) = 78 +12 + 48 = 138\ u$$
Question 5
1 / -0

The total number of oxygen atoms in $$0.2mol$$ of $${Na}_{2}{B}_{4}{O}_{7}.10{H}_{2}O$$ will be?

A
$$6.02\times {10}^{23}$$

B
$$1.02\times {10}^{24}$$

C
$$2.05\times {10}^{24}$$

D
$$2.05\times {10}^{23}$$

Solution

$$1\ mol$$ of $$Na_{2}B_{4}O_{4} \cdot 10H_{2}O$$ contains $$6.022\times 10^{23} molecules$$

$$0.2\ mol\ Na_{2}B_{4}O_{7}\cdot 10H_{2}O \rightarrow 0.2\times 6.022\times 10^{23} molecules$$

Then, the no. of oxygen atoms $$= 0.2\times 6.022\times 10^{23}\times 17\ atoms$$

$$= 2.047 \times 10^{24} molecules$$

Hence, Option (C) is correct.
Question 6
1 / -0

Number of electrons in 1.8 mL of $${H}_{2}O$$ are:

A
$$6.02 \times {10}^{23}$$

B
$$3.01 \times {10}^{24}$$

C
$$1.8 \times {10}^{23}$$

D
$$6.02 \times {10}^{25}$$

Solution

There is a total of 10 electrons in 1 $$ H_{2}O$$ Molecule (2 electrons from hydrogen and 8 electrons from oxygen.)

moles of $$ H_{2}O (n) = \dfrac{mass\, of\, substance}{motor \,mass}$$

18 g of $$H_2O$$ = 1 mole
mass of 1.8 ml of $$H_2O= 1.8 \ ml \times 1 \ g/ml= 1.8\ g$$ [$$\because Density \ of\ {H_2O} = 1 \ g/ml$$]

1.8 g of water = 0.1 moles of water

Therefore, 1.8 ml of $$ H_{2}O $$ means 0.1 mole of $$ H_{2}O$$

Number of molecules = n$$\times $$ Avogadro Number

Number of molecules = $$0.1 \times 6.022 \times 10^{23}$$

$$ = 0.6022 \times 10^{23}$$

Number of electron in total molecules of $$ H_{2}O$$ $$ = 0.6022 \times 10 ^{23} \times 10 $$

Number of electron in total molecules of $$ H_{2}O$$ $$ = 6.022 \times 10^{23}$$
Option $$A$$ is correct.
Question 7
1 / -0

$$10\ { dm }^{ 3 }$$ of $${ N }_{ 2 }$$ gas and $$10\ { dm }^{ 3 }$$ of gas X at the same temperature contain the same number of molecules. The gas X is

A
$$CO$$

B
$${ CO }_{ 2 }$$

C
$${ H }_{ 2 }$$

D
$$NO$$

Solution

$$CO$$

$$C-12,O-16$$

$$12+16=28$$

molar mass of $$N_2$$

$$N_2=14+4=28$$

Hence $$CO$$ is the answer
Question 8
1 / -0

Avogadro's number of helium atoms have a mass of

A
$$6.023\times 10^{23}g$$

B
$$4g$$

C
$$8g$$

D
$$4 \times\ 6.02 \times 10^{23}\ g$$

Solution

Here, Avogadro number of helium atoms means, $$6.023 \times 10^{23}$$ atoms of He.
As $$6.023\times 10^{23}$$ atoms equals to $$1$$ mole of Helium.
$$1$$ mole of Helium has mass $$4\ g$$
Hence, option $$B$$ is correct.
Question 9
1 / -0

If the atomic weight of oxygen were taken as $$90$$, then what would be the molecular weight of water?

A
$$101.25$$

B
$$104.52$$

C
$$112.5$$

D
$$142.5$$

Solution

If the atomic weight of oxygen = $$90\ g$$.
Then the factor by which oxygens atomic mass changes is, $$\cfrac{90}{16}= 5.62$$
If we go with the same assumption then the mass of hydrogen will also get change by the same factor. So, a new mass of two hydrogen atoms will be,
$$5.62\times 2=11.24 g$$

now, a total mass of water would be, $$90+11.24=101.25\ g$$

Hence, the correct option is (A).
Question 10
1 / -0

$$AB_2$$ and $$A_2B_2$$ are two compounds of the elements A and B. 0.25 mole of each of these compounds weights 9 g and 16 g respectively. Find the atomic masses of A and B.

A
28,4

B
28,8

C
13, 19

D
52,36

Solution

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Re-Attempt Weekly Quiz Competition

Atoms and Molecules Test - 62

Result

Atoms and Molecules Test - 62

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SHARING IS CARING

Question 1

1 : 8

1 : 1

7 : 64

1 :2

Solution

Question 2

Decrease twice

Increase two fold

Remain unchanged

Be a function of the molecular mass of the substance

Solution

Question 3

154

119

165

142

Solution

Question 4

128 u

112 u

116 u

138 u

90 u

Solution

Question 5

$$6.02\times {10}^{23}$$

$$1.02\times {10}^{24}$$

$$2.05\times {10}^{24}$$

$$2.05\times {10}^{23}$$

Solution

Question 6

$$6.02 \times {10}^{23}$$

$$3.01 \times {10}^{24}$$

$$1.8 \times {10}^{23}$$

$$6.02 \times {10}^{25}$$

Solution

Question 7

$$CO$$

$${ CO }_{ 2 }$$

$${ H }_{ 2 }$$

$$NO$$

Solution

Question 8

$$6.023\times 10^{23}g$$

$$4g$$

$$8g$$

$$4 \times\ 6.02 \times 10^{23}\ g$$

Solution

Question 9

$$101.25$$

$$104.52$$

$$112.5$$

$$142.5$$

Solution

Question 10

28,4

28,8

13, 19

52,36

Solution

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