Atoms and Molecules Test

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Atoms and Molecules Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following cations does not form soluble complex with excess NaOH solution as well as excess $$NH_4OH$$ solution ?

A
$$Cu^{+2}$$

B
$$Al^{+3}$$

C
$$Fe^{+3}$$

D
None of these
Question 2
1 / -0

Molecular mass of one mole of $$H_{2}SO_{4}$$ is ______.

A
$$98 a.m.u.$$

B
$$1.627\times 10^{-22}g$$

C
Both $$A$$ & $$B$$

D
$$1.627\times 10^{-25}g$$

Solution

$$Molecular \ mass = mass \ of \ hydrogen \ atom \times number \ of \ atoms + mass \ of \ sulfur \ atom \times number \ of \ atoms + mass \ of \ oxygen \ atom \times number \ of \ atoms$$
Molecular mass of $$H_2SO_4 = 1 \times 2 + 32 \times 1 + 16 \times 4 = 98 \ a.m.u.$$
Option A is correct.
Question 3
1 / -0

Hydrogen combines with oxygen to form $$H_2O$$ in which 16 g of oxygen combine with 2 g of hydrogen. Hydrogen also combines with carbon to form $$CH_4$$ in which 2 g of hydrogen combine with 6 g of carbon.If carbon and oxygen combine together then they will do show in the ratio of

A
13 : 32

B
6 : 16

C
1 : 2

D
12 : 24

Solution
Question 4
1 / -0

The mass of one molecule of carbon dioxide is:

A
$$26.49\times { 10 }^{ 24 }g$$

B
$$44\ g$$

C
$$7.30\times { 10 }^{ -23 }g$$

D
$$22\ g$$

Solution

As we know,

No. of molecules = $$No.\ of\ moles\times Avogadro\ Number$$

No. of moles $$ = \dfrac{No.\ of\ molecules} {Avogadro\ number}$$

Atomic Mass of $$C= 12u$$

Atomic Mass of $$O = 16u$$

Molecular Mass of $$CO_2=12+(2×16) =44\ u$$

Mass of One molecule of $$CO_2= \dfrac{44} {6.023\times 10^{23}}$$

= $$7.30×10^{-23}g$$

So, the correct option is $$C$$
Question 5
1 / -0

How many moles of $$CO_{ 2 }$$ are present in 220 mg ?

A
5 moles

B
0.005 mole

C
5000 moles

D
10 moles

Solution

Weight of given sample of $$CO_2=220mg=220\times 10^{-3}g$$

$$\text{no. of moles (n) }=\dfrac{\text{Weight of given sample}}{\text{molecular weight}}$$

$$=\dfrac{220\times 10^{-3}}{44}$$

$$=5\times 10^{-3}=0.005\,mole$$
Question 6
1 / -0

Atomic radius is measured in?

A
Centimetres (cm)

B
Nanometres (nm)

C
Metres (m)

D
Kilometres (Km)

Solution

Since atoms are too tiny, we cannot measure their radius in cm, m or km. Thus, they are measured in nanometres (nm).
Question 7
1 / -0

If the mass of neutrons is doubled and that of protons is halved, the molecular mass of $$H_2O$$ containing only $$H^1$$ and $$O^{16}$$ atoms will _____________.

A
increase by about $$25\%$$

B
decrease by about $$25\%$$

C
increase by about $$16\%$$

D
decrease by about $$16\%$$

Solution
Question 8
1 / -0

At $$25^oC$$, $$K_{sp}$$ value of $$AgCl$$ in water is $$1.8 \times 10^{-10}$$. If $$10^{-5}$$ moles of $$Ag^+$$ are added to this solution then $$K_{sp}$$ will be

A
$$3.13\times 10^{-10}$$

B
$$1.8 \times 10^{-10}$$

C
$$5.04 \times 10^{-5}$$

D
$$1.8 \times 10^{-8}$$
Question 9
1 / -0

Number of atoms in 560 g of Fe (atomic mass = 56 u) is

A
Twice that of 70 g N

B
Half that of 20 g H

C
Equal to that of 70 g N

D
None of these

Solution

Given, mass of Fe $$= 560\ g$$
Atomic mass of Fe $$= 56\ u$$

So, molar mass of Fe $$= 56\ gmol^{-1}$$

Number of moles $$= \dfrac {Mass}{Molar\ Mass}$$ $$= \dfrac {560}{56}$$ $$= 10$$

Number of atoms in 560 g Fe $$= Number\ of\ moles \times Avogadro\ number$$ $$= 10 \times 6.022 \times 10^{23}$$

Next, let's find the number of atoms in 70 g N
Given, mass of N $$= 70\ g$$

Atomic mass of N$$= 14\ u$$

So, molar mass of N $$= 14\ gmol^{-1}$$

Number of moles $$= \dfrac {Mass}{Molar\ Mass}$$ $$= \dfrac {70}{14}$$ $$= 5$$

Number of atoms in 70 g N$$= Number\ of\ moles \times Avogadro\ number$$ $$= 5 \times 6.022 \times 10^{23}$$

Comparing number of atoms in 560 g Fe and 70 g N:

Number of atoms in 560 g Fe $$= 10 \times 6.022 \times 10^{23}$$
Number of atoms in 70 g N $$= 5 \times 6.022 \times 10^{23}$$

So, number of atoms in 560 g Fe = Twice the number of atoms in 70 g N
Question 10
1 / -0

The number of moles in $$ 2 \times 10^{24} $$ atoms of iron are

A
3.3

B
4.5

C
5.2

D
2.1

Solution

Given, number of atoms of iron $$= 2 \times 10^{24}$$

We know, number of moles $$= \dfrac {Given\ number\ of\ particles}{Avogadro\ number}$$ $$= \dfrac {2 \times 10^{24}} {6.022 \times 10^{23}}$$ $$=3.3$$

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Atoms and Molecules Test - 72

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Atoms and Molecules Test - 72

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SHARING IS CARING

Question 1

$$Cu^{+2}$$

$$Al^{+3}$$

$$Fe^{+3}$$

None of these

Question 2

$$98 a.m.u.$$

$$1.627\times 10^{-22}g$$

Both $$A$$ & $$B$$

$$1.627\times 10^{-25}g$$

Solution

Question 3

13 : 32

6 : 16

1 : 2

12 : 24

Solution

Question 4

$$26.49\times { 10 }^{ 24 }g$$

$$44\ g$$

$$7.30\times { 10 }^{ -23 }g$$

$$22\ g$$

Solution

Question 5

5 moles

0.005 mole

5000 moles

10 moles

Solution

Question 6

Centimetres (cm)

Nanometres (nm)

Metres (m)

Kilometres (Km)

Solution

Question 7

increase by about $$25\%$$

decrease by about $$25\%$$

increase by about $$16\%$$

decrease by about $$16\%$$

Solution

Question 8

$$3.13\times 10^{-10}$$

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

$$5.04 \times 10^{-5}$$

$$1.8 \times 10^{-8}$$

Question 9

Twice that of 70 g N

Half that of 20 g H

Equal to that of 70 g N

None of these

Solution

Question 10

3.3

4.5

5.2

2.1

Solution

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