Atoms and Molecules Test

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Atoms and Molecules Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

Specific volume of cylindrical virus particle is $$6.02 \times {10^{ - 2}}\,\,cc/gm$$ whose radius and length are 7 A and 10 A respectively. If $${N_A} = 6.02 \times {10^{23}}.$$ Find molecular weight of virus.

A
$$15.4\,\,kg/mol$$

B
$$1.54 \times {10^4}\,\,kg/mol$$

C
$$3.08 \times {10^4}\,\,kg/mol$$

D
$$3.08 \times {10^3}\,kg/mol$$

Solution
Question 2
1 / -0

The number of electrons in $$3.1\ mg\ NO_{3}^{-}$$ is ______.
$$(Given: N_{A}=6\times 10^{23})$$

A
$$1.6\times 10^{-3}$$

B
$$9.6 \times 10^{20}$$

C
$$32$$

D
$$9.6 \times 10^{23}$$

Solution

Given that
Mass of $$NO_{3}^{-}, W_{NO_{3}^{-}} = 3.1\ mg = 3.1\times 10^{-3} g$$
Molar mass of $$NO_{3}^{-}, M_{NO_{3}^{-}} = 62\ g/mol$$

$$\therefore$$ Moles of $$NO_{3}^{-}, n_{NO_{3}^{-}} = \dfrac {W_{NO_{3}^{-}}}{M_{NO_{3}^{-}}}$$

$$= \dfrac {3.1\times 10^{-3} g}{62g/mol} = 5\times 10^{-5} mol$$
$$1\ NO_{3}^{-}$$ contain $$32$$ electrons.
$$\therefore$$ Number of electron $$(N_{e}) = 32\times n_{NO_{3}^{-}} \times N_{A}$$
$$= 32 \times (5\times 10^{-5}) \times 6.022\times 10^{23}$$
$$= 9.6352 \times 10^{20}$$
$$= 9.6\times 10^{20} electrons$$
Hence, option B is correct.
Question 3
1 / -0

Arrange the following in order of decreasing masses:

A
$$10 ^ { 23 }$$ molecules of $$\mathrm { CO } _ { 2 }$$ gas

B
0.1 g atom of silver

C
I gram of carbon

D
0.1 mole of $$\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$$

E
$$10 ^ { 23 }$$ atoms of calcium.
Question 4
1 / -0

The number of moles in $$ 1.6 \times 10^4 $$ grams of $$ Ca(OH)_2 $$ are

A
2.162

B
21.62

C
0.2162

D
216.2
Question 5
1 / -0

The present atomic weight scale is based on :-

A
$$^{12}{ C }$$

B
$$^{16}{ O }$$

C
$$^{1}{ H }$$

D
$$^{13}{ C }$$

Solution

The modern atomic weight scale is based on $$C^{12}$$. All the masses are measured relative to $$\text{carbon-12}$$. It is the standard unit for expressing the mass of an atom in amu (atomic mass unit). It is equal to the $$\frac {1}{12}$$ of the mass of the atom of carbon-12.
Option $$A$$ is correct.
Question 6
1 / -0

The weight of sulphuric acid needed for dissolving $$3 \mathrm { gm }$$ magnesium carbonate is:

A
$$3.5 \mathrm { gm }$$

B
$$7.0 \mathrm { gm }$$

C
$$1.7 \mathrm { gm }$$

D
$$17.0 \mathrm { gm }$$

Solution

The balanced equation is:
$$H_2SO_4 + MgCO_3 \rightarrow MgSO_4 + H_2CO_3$$
$$98\ g$$ $$84 \ g$$
Since, $$84 \ g \ MgCO_3 \ requires = 98 g \ H_2SO_4$$

Therefore, $$3\ g$$ of $$MgCO_3$$ = $$\dfrac{98\times3}{84} = 3.5 \ g$$
Question 7
1 / -0

The relative formula mass $$(M_{r})$$ of calcium carbonate $$(CaCO_{3})$$ is $$100$$. What is the mass of carbon present in $$100\ g$$ of calcium carbonate?

A
$$12\ g$$

B
$$36\ g$$

C
$$40\ g$$

D
$$60\ g$$

Solution

Solution:- (A) $$12 \ g$$
The relative formula masses $$\left( {M}_{r} \right)$$ of a compound is the sum of the relative atomic masses of the atoms in the numbers shown in the formula.
Mass of $$1$$ mole of carbon $$= 12 \; g$$
As $$CaC{O}_{3}$$ contains only $$1$$ mole of carbon, thus the mass of carbon present in $$100 \; g$$ calcium carbonate, i.e., $$1$$ mole $$CaC{O}_{3}$$ is $$12 \; g$$.
Question 8
1 / -0

A compound contains $$7$$ carbon atoms, $$2$$ oxygen atoms and $$9.96\times 10^{-24}\,g$$ of other elements. The molecular mass of compound is $$(N_A =6\times 10^{23})$$

A
$$122$$

B
$$116$$

C
$$148$$

D
$$154$$

Solution

Molecular mass $$=(7\times 12)+(2\times 16)+(9.96\times 10^{-24})\times 6.02\times 10^{23}$$
$$=84+32+6$$
$$=$$ $$122\ gm$$
option $$(a)$$
Question 9
1 / -0

The volume of one mole of water at $$277\ K$$ is $$18\ ml$$. One $$ml$$ of water contains $$20$$ drops. The number of molecules in one drop of water will be $$(N_{A}=6\times 10^{23})$$

A
$$1.07\times 10^{21}$$

B
$$1.67\times 10^{21}$$

C
$$2.67\times 10^{21}$$

D
$$1.67\times 10^{20}$$

Solution

$$1$$ mole $$=185\ ml$$ (given)
$$1\ ml=20\ drops$$

$$\therefore 1$$ mole $$=18\times 20$$
$$=3360\ drops$$.

$$1$$ mole $$=6.022\times 10^{23}$$ molecules

$$\Rrightarrow 360\ drops=6.022\times 10^{23}$$ molecules

$$\therefore 1\ drops=x$$ molecules
$$360\times x=1\times 6.022\times 10^{23}$$
$$x=1.67\times 10^{21}$$ molecules

Hence, Option $$B$$ is the correct answer.
Question 10
1 / -0

For $$CuSO_4.5H_2O,$$ which is the correct mole relationship?

A
mole of $$Cu= 9 \times mole\ of\ O$$

B
mole of $$Cu= 4 \times mole\ of\ O$$

C
mole of $$Cu= 5 \times mole\ of\ O$$

D
mole of $$O= 9 \times mole\ of\ Cu$$

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Atoms and Molecules Test - 64

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Atoms and Molecules Test - 64

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Question 1

$$15.4\,\,kg/mol$$

$$1.54 \times {10^4}\,\,kg/mol$$

$$3.08 \times {10^4}\,\,kg/mol$$

$$3.08 \times {10^3}\,kg/mol$$

Solution

Question 2

$$1.6\times 10^{-3}$$

$$9.6 \times 10^{20}$$

$$32$$

$$9.6 \times 10^{23}$$

Solution

Question 3

$$10 ^ { 23 }$$ molecules of $$\mathrm { CO } _ { 2 }$$ gas

0.1 g atom of silver

I gram of carbon

0.1 mole of $$\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$$

$$10 ^ { 23 }$$ atoms of calcium.

Question 4

2.162

21.62

0.2162

216.2

Question 5

$$^{12}{ C }$$

$$^{16}{ O }$$

$$^{1}{ H }$$

$$^{13}{ C }$$

Solution

Question 6

$$3.5 \mathrm { gm }$$

$$7.0 \mathrm { gm }$$

$$1.7 \mathrm { gm }$$

$$17.0 \mathrm { gm }$$

Solution

Question 7

$$12\ g$$

$$36\ g$$

$$40\ g$$

$$60\ g$$

Solution

Question 8

$$122$$

$$116$$

$$148$$

$$154$$

Solution

Question 9

$$1.07\times 10^{21}$$

$$1.67\times 10^{21}$$

$$2.67\times 10^{21}$$

$$1.67\times 10^{20}$$

Solution

Question 10

mole of $$Cu= 9 \times mole\ of\ O$$

mole of $$Cu= 4 \times mole\ of\ O$$

mole of $$Cu= 5 \times mole\ of\ O$$

mole of $$O= 9 \times mole\ of\ Cu$$

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