Atoms and Molecules Test

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Atoms and Molecules Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

How many moles of magnesium phosphate, $$Mg_3(PO_4)_2$$ will contain 0.25 mole of oxygen atoms ?

A
$$0.02$$

B
$$3.125 \times 10^{-2}$$

C
$$1.25 \times 10^{-2}$$

D
$$2.5 \times 10^{-2}$$

Solution

$$1$$ molecule of $$Mg_3(PO_4)_2$$ has $$8$$ atoms of Oxygen.

$$\Rightarrow 1$$ mole of $$Mg_3(PO_4)_2$$ has $$8$$ mole of Oxygen atoms.

Now, it is given $$0.25$$ mole of oxygen atoms
$$\therefore$$ Moles of $$Mg_3(PO_4)_2$$ having $$0.25$$ mole of $$O$$ atoms is $$\cfrac {0.25\times 1}{8}$$ moles
$$=0.03125$$ moles

$$=3.125\times 10^{-2}$$ moles

Hence, option B is correct.
Question 2
1 / -0

Number of atoms in 558.5 g of Fe:

(Atomic mass of Fe = 55.85 u)

A
$$15.09 \times 10^{25}$$

B
$$6.023 \times 10^{24}$$

C
$$38$$

D
$$ 5 \times 2\times 6.023 \times 10^{23}$$

Solution

Given mass of Fe = $$ 558.5\ g$$

Molar mass of Fe = $$ 55.85\ g/mol$$ (As we know that atomic mass and molar mass are numerically equal, only units change, and the atomic mass is given in question)

No. of moles of Fe = $$\dfrac {Given\ mass}{Molar\ mass}$$ = $$\dfrac {558.5}{55.85}$$ = $$10\ moles$$

Hence, no. of atoms in 10 moles of Fe = $$ 10 \times Avogadro\ number$$ = $$10 \times 6.023 \times 10^{23}$$ = $$6.023 \times 10^{24}\ atoms$$
Question 3
1 / -0

The ratio of unpaired electrons present in d orbitals of $$Co^{2+}$$ and $$Cr^{3+}$$ is ____.

A
2:1

B
1:3

C
3:1

D
4:3

Solution

Ground state configuration of cobalt
$$=[Ar]3d^74s^2$$
Ground state configuration of $${ Co }^{ 2+ }$$
Remove $$2$$ electrons
So only one unpaired $${ e }^{ - }$$
Ground configuration of $$Cr=4{ s }^{ 1 }3{ d }^{ 5 }\\ { Cr }^{ 3+ }=4{ s }^{ 1 }3{ d }^{ 2 }$$
unpaired electrons $$=3$$
$$\boxed { { Co }^{ 2+ }:{ Cr }^{ 3+ }=1:3 } $$
Question 4
1 / -0

Specific volume of cylindrical virus particles is $${ 6.02\quad \times }{ 10 }^{ -2 }$$ cc/gm whose radius and length are 7 $$\mathring A$$ and 10 $$\mathring A$$, respectively. If $${ N }_{ A }=6.02\times\ { 10 }^{ 23 }$$ find molecular weight of virus.

A
$$15.4 $$ kg/mol

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

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

D
$$1.54\quad \times \quad { 10 }^{ 3 }$$ kg/mol

Solution

Solution:- (A) $$15.4 \; {Kg}/{mol}$$

Volume of cylinder $$= \pi {r}^{2}h$$

As the virus is cylindrical.

$$\therefore$$ Volume of one virus $$=$$ volume of cylinder

As the radius and length of virus are $$7 \mathring{A}$$ and $$10 \mathring{A}$$ respectively.

$$\therefore$$ Volume of $$1$$ virus $$= \cfrac{22}{7} \times {7}^{2} \times 10 = 1540 \times {10}^{-24} {cm}^{3} = 1.54 \times {10}^{-21} {cm}^{3}$$ $$[\because \mathring 1 = 10^{-8} cm^3]$$

Volume of $$1$$ mole of virus $$= {N}_{A} \times $$volume of $$1$$ virus

$$\Rightarrow$$ Volume of $$1$$ mole of virus $$= 6.02 \times {10}^{23} \times 1.54 \times {10}^{-21} = 6.02 \times 1.54 \times {10}^{2} {cm}^{3}$$

Given that specific volume of virus is $$6.02 \times {10}^{-2} {{cm}^{3}}/{g}$$

$$\therefore \text{Molecular weight} = \cfrac{\text{volume of 1 mole}}{\text{Specific volume}} = \cfrac{6.02 \times 1.54 \times {10}^{2}}{6.02 \times {10}^{-2}} = 1.54 \times {10}^{4} \; {g}/{mol} = 15.4 \; {Kg}/{mol}$$

Hence, the correct option is A.
Question 5
1 / -0

If 10$$^{21}$$ molecules are removed from 200 mg of CO$$_{2}$$, the number of moles of CO$$_{2}$$ left is :

A
$$2.88 \times 10^{-3}$$

B
$$28.8 \times 10^{-3}$$

C
$$0.288 \times 10^{-3}$$

D
$$1.66 \times 10^{-3}$$

Solution

Answer:-
Molar mass of $$C{O}_{2}$$ = 44g

Given mass = 200 mg = 0.2g

Number of moles = $$\cfrac{0.2}{44} = \cfrac{1}{220}$$

$$\text{Number of moles} = \cfrac{\text{number of molecules}}{\text{Avogadro's number}}$$

$$\Rightarrow \text{Number of molecules} = \text{Avogadro's number} \times \text{number of moles}$$

$$\Rightarrow \text{No. of molecules} = 6.022 \times {10}^{23} \times \cfrac{1}{220} = 2.73 \times {10}^{21} \text{ Molecules in } \cfrac{1}{220} moles$$

As $${10}^{21}$$ molecules are removed,
$$\therefore$$ No. of molecules left = $$2.73 \times {10}^{21} - {10}^{21} = 1.73 \times {10}^{21}$$

$$\therefore \text{ No. of moles} = \cfrac{\text{No. of molecules}}{\text{Avogadro's number}} = \cfrac{1.73 \times {10}^{21}}{6.23 \times {10}^{23}} $$

$$= 2.88 \times {10}^{-3}$$

Hence, option $$A$$ is correct.
Question 6
1 / -0

If we assume that one-sixth the mass of an atom of $$^{12}C$$ isotope is taken as the reference, the mass of one molecule of oxygen will:

A
be double its original value

B
be half its original value

C
be the same

D
increase by four fold

Solution

We know that $$12\ g\ of\ Carbon=6.023\times \ 10^{23} atoms$$

So $$1\ g=\frac{6.023\ X\ 10^{23}}{12}$$

$$6\ g\ of \ carbon=6\times \frac{6.023\ X\ 10^{23}}{12}$$

or

$$=\frac{1}{2}\times 6.023\ \times 10^{23}$$

This is our new Avogadro's number according to the given conditions.

Or
$$1\ atomic\ mass\ unit=\frac{1}{N_A}=\frac{2}{6.023\ X\ 10^{23}}$$
Mass of $$1$$ mole of $$O_2=32\times \ Avogrado's\ number\times \frac{2}{6.023\ \times 10^{23}}$$
$$=32\times \frac{6.023\ \times 10^{23}}{2}\times \frac{2}{6.023\ \times 10^{23}}$$
$$=32g$$
Hence, there is no change in the mass number of $$O_2.$$
Option $$\text{C}$$ is correct.
Question 7
1 / -0

How many oxygen atoms will be present in $$88\ g$$ of $$CO_2$$?

A
$$24.08 \times 10^{23}$$

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

C
$$44 \times 10^{23}$$

D
$$22 \times 10^{24}$$

Solution

1 mol of $$CO_2=12+32=44\ g$$
Number of oxygen in 1 molecule of $$CO_2=2$$

$$\therefore$$ 44 g $$CO_2$$ contains 2 mol oxygen

$$88\ g \ CO_2$$ contains $$=\cfrac{2}{44} \times 88=4$$ mol oxygen

as, $$1 \ mol =6.022 \times 10^{23}$$ atoms

$$4\ mol\ oxygen =4 \times 6.022 \times 10^{23}$$ oxygen atoms

$$=24.08 \times 10^{23}$$ oxtgen atoms

Hence, option $$A$$ is correct.
Question 8
1 / -0

How many grams of $$CaO$$ are required to react with $$852\ g$$ of $$P_4O_{10}$$ according to below given reaction:
$$6CaO+P_4O_{10}\to 2Ca_3(PO_4)_2$$

A
852 g

B
1008 g

C
85 g

D
7095 g

Solution

The balanced reaction is :
$$6CaO + P_4O_{10} \rightarrow 2Ca_3 (PO_4)_2$$

Molar mass of $$P_4O_{10}=4 \times 31 + 10 \times 16=284$$ g

Moles of $$P_4O_{10} =\frac{mass}{molar\ mass}=\frac{ 852 }{284} = 3$$ moles

1 mol of $$P_4O_{10}$$ react with 6 mol of $$CaO$$
3 moles will react with$$= 3 \times 6 = 18$$ mol of $$CaO$$
Mass of 1 mol of $$CaO=40+16=56$$ g
Mass of 18 moles of $$CaO= 18 \times 56 = 1008$$ g
option B is correct.
Question 9
1 / -0

P and Q are two elements which form $${ P }_{ 2 }{ Q }_{ 3 }$$ and $${ PQ }_{ 2 }$$. If 0.15 mole of $${ P }_{ 2 }{ Q }_{ 3 }$$ weight 15.9 g and 0.15mole of $${ PQ }_{ 2 }$$ weight 9.3 g. what are atomic weights of P and Q respectively?

A
18 and 26

B
26 and 26

C
26 and 18

D
18 and 18

Solution

Using- no. of mole $$=\cfrac{\text {weight}}{\text{molecular weight}}$$

$$P_2Q_3 \Rightarrow $$ $$2P + 3Q =15.9/0.15= 106$$....(1)

$$PQ_2 \Rightarrow $$ $$P + 2Q = 9.3/0.15=62$$........(2)

On solving eqn. (1) & (2) we get-

$$P =26, Q = 18$$
Question 10
1 / -0

Who first gave the concept of 'atom'?

A
John Dalton

B
Aristotle

C
Kanada

D
Kapila

Solution

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Atoms and Molecules Test - 70

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Atoms and Molecules Test - 70

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Question 1

$$0.02$$

$$3.125 \times 10^{-2}$$

$$1.25 \times 10^{-2}$$

$$2.5 \times 10^{-2}$$

Solution

Question 2

$$15.09 \times 10^{25}$$

$$6.023 \times 10^{24}$$

$$38$$

$$ 5 \times 2\times 6.023 \times 10^{23}$$

Solution

Question 3

2:1

1:3

3:1

4:3

Solution

Question 4

$$15.4 $$ kg/mol

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

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

$$1.54\quad \times \quad { 10 }^{ 3 }$$ kg/mol

Solution

Question 5

$$2.88 \times 10^{-3}$$

$$28.8 \times 10^{-3}$$

$$0.288 \times 10^{-3}$$

$$1.66 \times 10^{-3}$$

Solution

Question 6

be double its original value

be half its original value

be the same

increase by four fold

Solution

Question 7

$$24.08 \times 10^{23}$$

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

$$44 \times 10^{23}$$

$$22 \times 10^{24}$$

Solution

Question 8

852 g

1008 g

85 g

7095 g

Solution

Question 9

18 and 26

26 and 26

26 and 18

18 and 18

Solution

Question 10

John Dalton

Aristotle

Kanada

Kapila

Solution

User

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