Tag: discovery of subatomic particles

Questions Related to discovery of subatomic particles

The ratio of mass of proton to electron is about:

  1. $1.8\times 10^{-3}$

  2. $1.8\times 10^{3}$

  3. 1

  4. 0


Correct Option: B
Explanation:

Mass of proton = $1.6\times 10^{-27}$kg and mass of electron = $9.1\times 10^{-31}$kg
$\therefore\dfrac{\text{mass of proton}}{\text{mass of electron}} = \dfrac{1.6\times 10^{-27}}{9.1\times 10^{-31}} = 1.8\times 10^{3}$

The electronic configuration of an ion $M^{2+}$ is 2, 8, 14. If its mass 56, the number of neutrons in its nucleus is :

  1. 30

  2. 32

  3. 34

  4. 42


Correct Option: A
Explanation:

$M^{2+}=2, 8, 14;\Rightarrow $ Electronic Configuration of $M=2, 8, 16$
So. total electrons in M will be $=26$
i.e., atomic number $(Z)=26$, Mass number (given) $=56$
$\therefore$ Number of neutrons$=A-Z=56-26=30$

Hence, the correct option is A.

Suppose the chemists hed selected $10^{20}$ as the number of particles in a mole. The molar mass of oxygen gas would be (Use Avogadro number $=6.0\times 10^{23}$)

  1. $5.33\times 10^{-3}g$

  2. $5.35\times 10^{-23}g$

  3. $5.33\times 10^{-43}g$

  4. $32\times 10^{3}g$


Correct Option: A
Explanation:

Molar mass of oxygen we know is $32$g.

$\therefore 1$ molecule oxygen weighs $\cfrac { 32 }{ 6\times { 10 }^{ 23 } } $g
$\Rightarrow { 10 }^{ 20 }$ molecules would weigh $\cfrac { 32 }{ 6\times { 10 }^{ 23 } } \times { 10 }^{ 20 }$
$\therefore 1 $mole would weigh $=5.33\times { 10 }^{ -3 }$g.

The number of electrons present in $100ml$ of $0.1N$ ${H} _{2}{SO} _{4}$ is:

  1. $6.85\times {10}^{22}$

  2. $7.50\times {10}^{23}$

  3. $1.5\times {10}^{23}$

  4. $1.8\times {10}^{22}$


Correct Option: C
Explanation:

Normality=$n\times Molarity$

$\therefore 0.1=2\times \cfrac { moles\quad  of\quad  H _2SO _4}{ 0.1 }$
$\therefore Moles\quad  of\quad  H _2SO _4=0.005\quad moles$

$1$ molecule of $H _2SO _4$ contains $50$ electrons.

No. of molecules in $0.005$ moles=$0.005\times6.022\times {10}^{ 23 }$
=$0.03011\times {10}^{ 23 }$ molecules

$\therefore$ No. of electrons in $0.005$ moles=$0.03011\times {10}^{ 23 }\times 50$
=$1.5\times {10}^{ 23 }$ electrons

Total number of electrons present in $14\ gm$ of nitrogen gas is:

$[N _A = 6\times 10^{23}]$

  1. $6\times { 10 }^{ 23 }$

  2. $4.2\times { 10 }^{ 24 }$

  3. $3\times { 10 }^{ 23 }$

  4. $4.2\times { 10 }^{ 23 }$


Correct Option: B

The ratio between the number of neutrons present in $C^{12}$ and $Si^{30}$ atoms is:

  1. 3:8

  2. 2:5

  3. 3:7

  4. 1:1


Correct Option: A

The number of electrons present in  $3\mathrm { d }$  of  $\mathrm { Cu }  $ is :

  1. $20$

  2. $10$

  3. $16$

  4. $24$


Correct Option: B
Explanation:

Electronic configuration of copper is- $1s^22s^22p^63s^23p^64s^13d^{10}$

There are $10$ electrons in $3d-$ shell to attain fully filled configuration.

The number of electrons having $n+s=2.5$ in sulphide ion is :-

  1. $3$

  2. $4$

  3. $6$

  4. $8$


Correct Option: B

The approximate atomic weight of electron or positron per mol is:

  1. $5.48\times 10^{-6} g mol^{-1}$

  2. $ 5.48\times 10^{-8} g mol^{-1}$

  3. $5.48\times 10^{-4} g mol^{-1}$

  4. None of these


Correct Option: C
Explanation:

Mass of one electron $=9.1\times 10^{-27} g $

So, mass of one mole of electron $=m\times N _o$ $=5.48\times 10^{-4} g mol^{-1}$ as  $N _o = 6.023\times 10^{23}$.

Rest mass of an electron is $9.11\times 10^{-31}$ kg. Molar mass of the electron is:

  1. $1.50\times 10^{-31}kg$ $mol^{-1}$

  2. $9.11\times 10^{-31}kg$ $mol^{-1}$

  3. $5.5\times 10^{-7}kg$ $mol^{-1}$

  4. $6.02\times 10^{23}kg$ $mol^{-1}$


Correct Option: C
Explanation:

Rest mass of an electron is $9.11\times 10^{-31}kg$.
So, mass of one mole of electron is $9.11\times 10^{-31}\times6.023\times10^{23} = 5.5\times 10^{-7}kg/mole$