Chemistry · Physics

Atomic Structure and Mass

308 Questions

Atomic structure and mass focus on the composition of atoms, including protons, neutrons, electrons, and isotopes. This topic is essential for the chemistry sections in engineering and civil services competitive exams. Review these questions to practice calculating atomic mass, neutron count, and isotopic distributions.

Atomic mass unit calculationsNeutron and proton countsIsotopic distribution averagesLaw of triadsRest energy of atoms

Atomic Structure and Mass Questions

Multiple choice chemistry classification of elements- the periodic table development of the modern periodic table into the history mendeleef's periodic table

The atomic masses of Li and K are 7 and 39, respectively. According to law of triads the atomic mass of Na will be:

  1. 23

  2. 32

  3. 46

  4. 64
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Dobereiner arranged the elements in increasing order of atomic masses. He found that the atomic mass of the middle element was approximately equal to the arithmetic mean (average) of the atomic masses of the other two elements of that triad when they are arranged in their increasing order of atomic mass.

So Li=7,K=39

Atomic mass of Na = $\dfrac{7+39}{2}=\dfrac{46}{2}=23$
Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

$E _n$ and $J _n$ denote the total energy magnitude and the  angular momentum of an electron in the nth allowed orbit of the Both atom .Then:

  1. $E _n \pi J _n$
  2. $E _n \pi \frac {1} {J}$
  3. $E _n \pi { J } _{ n }^{ 2 }$
  4. $E _n \pi \frac {1} {J^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the Bohr model, E is proportional to 1/n^2 and J (angular momentum) is proportional to n. Therefore, E is proportional to 1/J^2.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Which term describes the mass of $6.022\times { 10 }^{ 23 }$ representative particles?

  1. Molar mass

  2. Avogadro's number

  3. Empirical formula

  4. Molecular formula
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molar mass of a compound describes the total mass of $6.023 \times 10^{23}$ atoms or particles of the compound.

Ex: Molar mass of $H _2 = 2\space g$

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Which of the following statements is the most accurate with regard to the significance of Avogadro's number, $6.02 \times 10^{23}$ ?

  1. It is the conversion factor between grams and atomic mass units.

  2. It is a universal physical constant just as the speed of light.
  3. It is the number of particles that is required to fill a $1\ L$ container.
  4. It is the inverse diameter of an $H$ atom .
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A carbon atom weighs 12 AMU, one AMU is equivalent to $1.66 \times 10^{-24}$ grams. Conversely, one gram is equivalent to $6.022 \times 10^{23}$ AMU, which is called Avogadro's number


so the answer is A

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Mass of the one atom of the element X is $1.66 \times 10^{-26}$ g. Number of atoms in 1 g of the element is:

  1. $1.66 \times 10^{-26}$
  2. $1.66 \times 10^{25}$
  3. $1.66 \times 10^{-24} x No$
  4. $6.024 \times 10^{25}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mass of one atom = $1.66 \times 10^{-26} g$
$1.66 \times 10^{-26} gm \rightarrow 1 \, atom$
$1 \, gm \rightarrow \dfrac{1}{1.66 \times 10^{-26}} $ atom
$= 6.024 \times 10^{25} atom$

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Avogadro's number is NOT equal to:

  1. the number of atoms in 11.2 L of $\displaystyle { O } _{ 2 }$ at STP
  2. the number of atoms in 1 mole of $\displaystyle He$ at STP

  3. the number of electrons in 96,500 coulombs
  4. the number of $\displaystyle { SO } _{ 4 }^{ 2- }$ ions in 1 L of 0.5 M sulphuric acid
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$Avogadro's \ Number= N _A= 6.02 \times 10^{23}$
$1$ mole of any substance has $N _A$ number of atoms, molecules or ions etc. 
Therefore, moles of $SO _4^{2-}$ ions in $1 \ L$ of $0.5 \ M \ H _2SO _4$ will be 
$Molarity= \cfrac {Number \ of \ moles}{Volume \ of \ solution \ (in \ L)}$
$0.5 = \cfrac {moles}1$
$Number \ of \ moles = 0.5$
Hence, because $1$ mole of a substance $= N _A$
$\therefore \ 0.5$ moles of $SO _4^{2-} \neq N _A $ (Avogadro Number)
Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Which of the following expressions is equal to the number of iron ($Fe$) atoms present in $10.0$ g $\displaystyle Fe$ ? (atomic mass of $\displaystyle Fe$ = $55.9$ amu) 

  1. $ 10\times 55.9\times( 6.022\times { 10 }^{ 23 }) $ atoms
  2. $\dfrac{( 6.022\times { 10 }^{ 23 })}{10}\times 55.9$ atoms
  3. $ 10\times \dfrac {( 6.022\times { 10 }^{ 23 } )}{ 55.9}$ atoms
  4. $\dfrac {55.9}{10} \times ( 6.022\times { 10 }^{ 23 } ) $
  5. $ \dfrac {10}{ ( 55.9\times 6.022\times { 10 }^{ 23 } )} $ atoms
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression $\displaystyle 10\times \left( 6.022\times { 10 }^{ 23 } \right)/ 55.9$ is equal to the number of iron (Fe) atoms present in 10.0 g Fe.
The atomic mass of Fe is 55.9 g/mol.
The mass of Fe is 10.0 g. Mass is divided with atomic mass to obtain number of moles.
The number of moles of Fe $ =  \dfrac {10.0}{55.9}$ moles.
The number of moles is multiplied with avogadro's number to obtain the number of Fe atoms.
The number of Fe atoms  $ =  \dfrac {10.0}{55.9} \times 6.023 \times 10^{23}$.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

1 g of $^{12}C$ contains $6.022 \times 10^{23}$ atoms of the isotope.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Because according to Avegadro's law,

$1$ mole of $C$ contains $6.022\times { 10 }^{ 23 }$ atoms and $1$ mole of $C$ weighs $12gm$
$\therefore$  $12g$ of $C$ weighs $6.022\times { 10 }^{ 23 }$ atoms

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Calculate the number of iron atoms in a piece of iron weighing $2.8 g$. (Atomic mass of iron $=56$)

  1. $30.11\times { 10 }^{ 23 }$ atoms
  2. $3.11\times { 10 }^{ 23 }$ atoms
  3. $3.0115\times { 10 }^{ 22 }$ atoms
  4. $301.1\times { 10 }^{ 23 }$ atoms
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that, weight of iron =2.8 g
Atomic mass= 56
moles of iron$= \dfrac{\text{wt. of iron}}{\text{atomic mass}}=\dfrac{2.8}{56}=0.05\ moles$
1 mole $= 6.022 \times 10^{23}\ atoms$
0.05 moles $= 6.022 \times 0.05 \times 10^{23}\ atoms=3.0115 \times 10^{22}$ atoms
Option C is correct.
Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

What would be the approximate weight of $1.204\times 10^{24}$ bromine atoms?

  1. 80 grams

  2. 120 grams

  3. 160 grams

  4. 180 grams

  5. 200 grams
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The atomic weight of bromine is 80.

So, 80-gram bromine contains Na atoms.
So, $1.204\times10^{24}$ bromine atoms will weigh:  $80\div 6.02\times10^{23}\times 1.204\times10^{24}=160\ gram$

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

What could be better than a dozen $(12)$ donuts? How about a baker's dozen $(13)$ of donuts? Another large unit of measurement is known as Avogadro's number $(6.022\times 10^{23})$.
What is TRUE about Avogadro's number?

  1. Avogadro's number is the number of particles in one mole of any element.
  2. The molar mass of a substance will contain $6.022\times 10^{23}$ molecules.
  3. There are $6.022\times 10^{23}g$ of carbon in $12$ mol of $C - 12$.
  4. $6.022\times 10^{23}$ atoms of any element will have the exact same mass regardless of the identity of the element.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Avagadro constant i.e. $6.023 \times 10^{23}$ particles is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by $1\space mole$. 

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

A student wrote the relation for one unified atomic mass unit (u) as $1u=931.5MeV$. What is the correct relation?

  1. $1 u\times c=931.5 MeV$
  2. $1 u\times c^2=931.5 MeV$
  3. $\dfrac{1u}{c^2}=931.5 MeV$
  4. $(1u)^2\times c=931.5Me V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1u=931.5 MeV$
we know $1u=1.66\times 10^{-21}kv$
$931.5 MeV=931.5\times 10^6\times 1.6\times 10^{-19}J=1.49\times 10^{-10}$
$E=(1u)(c^2)=(1.66\times 10^{-27})(3\times 10^8)^2J=1.494\times 10^{-10}J$
so $(1u)(c)^2=931.5MeV$