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Determination of atomic and isotopic mass - class-XI

Description: determination of atomic and isotopic mass
Number of Questions: 17
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Tags: chemistry some basic concepts of chemistry
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There are two isotopes of an element with atomic mass $z$. Heavier one has atomic mass $z+2$ and lighter one has $z-1$, then an abundance of lighter one is:

  1. $66.6\%$

  2. $96.7\%$

  3. $6.67\%$

  4. $33.3\%$


Correct Option: A
Explanation:

Abundance of an isotope can be calculated as :


Let abundance of isotope with atomic mass $Z+2$ is x
Then abundance of isotope $Z-1$ is $1-x$

$(Z+2)x+(Z-1)(1-x)\quad =Z$

On solving this, we get
$x=\dfrac { 1 }{ 3 } $
Therefore,% abundance of isotope with atomic no.$Z+2$ is =$=\dfrac { 1 }{ 3 } \times 100=33.33$%
Therefore,% abundance isotope with atomic no. $Z-1$ is =$100-33.33=67.66%$%
Therefore isotope with atomic $Z-1$ is higher in % abundance

$ _{ 17 }^{ 35 }{ Cl }$ and $ _{ 17 }^{ 37 }{ Cl }$ are two isotopes of chlorine. If average atomic mass is $35.5$ then ratio of these two isotopes is:

  1. $35 : 37$

  2. $1 : 3$

  3. $3 : 1$

  4. $2 : 1$


Correct Option: C
Explanation:

Solution:- (C) $3 : 1$

Let $x$ and $y$ be the fraction of ${ _{17}^{35}{Cl}}$ and ${ _{17}^{37}{Cl}}$ in ${ _{17}^{35.5}{Cl}}$.
Therefore,
Average at. mass $= \cfrac{x \times 35 + y \times 37}{x + y}$
$35.5 = \cfrac{35x + 37y}{x+y}$
$\Rightarrow 35.5 x + 35.5 y = 35x + 37y$
$\Rightarrow 35.5 x - 35x = 37 y - 35.5 y$
$\Rightarrow 0.5 x = 1.5 y$
$\Rightarrow \cfrac{x}{y} = \cfrac{1.5}{0.5} = \cfrac{3}{1}$
Hence the ratio of ${ _{17}^{35}{Cl}}$ and ${ _{17}^{37}{Cl}}$ in ${ _{17}^{35.5}{Cl}}$ is $3 : 1$.

If isotopic distribution of $ C-12 $ and $ C-14 $ is  98 %  and  2 %  respectively, what would be the number of $ C-14 $ isotope in $ 12 gm $  carbon sample?

  1. $ 1.032 \times 10^{22} $

  2. $ 3.01 \times 10^{23} $

  3. $ 5.88 \times 10^{23} $

  4. $ 6.02 \times 10^{23} $


Correct Option: A
Explanation:

1 mole of carbon contains $N _A$ molecules.


$14g$ of $C-14$ contains $6.023 \times 10^{23}$ atoms

$1g$ of $C-14= \dfrac{6.023 \times 10^{23}}{14}$


$12g$ of $C-14$ atom contains $=\cfrac {6.023 \times 106{23}\times 12}{14}$

                                                  $=5.16 \times 10^{23}$ atoms

Now $2$% of $5.16 \times 10^{23}=\cfrac {5.16 \times 10^{23}\times 2}{100}$

                                         $=1.032 \times 10^{22}$ atoms .

Chlorine has two naturally occurring isotopes, $^{35}Cl$ and $^{37}Cl$. If the atomic mass of Cl is 35.5 the ratio of natural abundance of $^{35}Cl$ and $^{37}Cl$ is closest to :

  1. $ 3 : 5 $

  2. $3 : 1$

  3. $2 : 5 $

  4. $4 : 1$


Correct Option: B

A sample of perfect gas that initially occupies $15.0L$ at $300K$ and $1.0$ bar is compressed isothermally. To what volume must the gas be compressed to reduce its entropy by $5.0J/K$? $\left[ \ln { 0.36 } =-1.0,\ln { 2.7 } =1.0 \right] $

  1. $5.4L$

  2. $8.22L$

  3. $40.5L$

  4. $5.56L$


Correct Option: A

If an element $z$ having atomic weight $x$ exists in two isotopes of mass number $(x-1)$ and $(x+2)$ then, the percentage abundance of heavier isotopes is:    

  1. $25\%$

  2. $66.6\%$

  3. $33.3\%$

  4. $20\%$


Correct Option: B

A system undergoes a process in which the entropy change is $+5.51J{K}^{-1}$. During the process, $1.50kJ$ of heat is added to the system at $300K$. The correct information regarding the process is

  1. the process is thermodynamically reversible

  2. the process is thermodynamically irreversible

  3. the process may or may not be thermodynamically reversible

  4. the process must be isobaric


Correct Option: B

The table shows the numbers of particles present in the nuclei of four atoms or ions.

protons neutrons electronic structure
$1$ $18$ $22$ $2, 8, 8$
$2$ $19$ $20$ $2, 8, 8$
$3$ $19$ $21$ $2, 8, 8, 1$
$4$ $20$ $20$ $2, 8, 8, 2$

Which two particles belong to the same element?

  1. $1$ and $2$

  2. $1$ and $4$

  3. $2$ and $3$

  4. $2$ and $4$


Correct Option: C
Explanation:

$\text{An element always consist same number of proton and electrons.}$

$\text{Number of neutrons can be changed in an element based on its isotrops.}$
$\text{Option C is correct.}$

The average atomic mass of copper is $63.546$ amu .Natural copper consists of two iostopes: $^{63} Cu$ and $^{65} Cu$.Their natural abundances are $69.09\%$ and $30.91\%$ respectively. If the mass of $^{63} Cu$ isotope is $62.9298$ amu ,What is the mass of $^{65} Cu$ isotope?

  1. 64.9000

  2. 65.1233

  3. 64.9233

  4. 65.1933


Correct Option: C

Europium consists of $47.8$% Eu-$151$ and $52.2$% Eu-$153$. Calculate average atomic mass?

  1. $152.2$

  2. $152.1$

  3. $151.9$

  4. $152$


Correct Option: D
Explanation:

Average atomic mass is given by:


$M _{av}=\dfrac{151\times0.478+153\times0.522}{0.478+0.522}=152.044\ gm/mol$

 Rhenium (Re) consists of $37.1$% $185$ Re and $62.9$% $187$ Re. Calculate the relative atomic mass?

  1. $185.6$

  2. $185.9$

  3. $186.3$

  4. $186.1$


Correct Option: C

The relative abundance of two isotopes of atomic masses 85 and 87 are 75% and 25% respectively. The average atom mass of element is: 

  1. 86

  2. 40

  3. 85.5

  4. 75.5


Correct Option: C

Antimony (Sb) is $57.2$% Sb-$121$ and $42.8$% Sb-$123$. Calculate the average atomic mass?

  1. $121.6$

  2. $120.8$

  3. $122.4$

  4. $121.9$


Correct Option: D
Explanation:

Average Atomic mass is given as:


$M _{av}=\dfrac{121\times0.572+123\times0.428}{1}=121.856\ gm/mol$

The relative atomic mass of an atom is:

  1. measured in atomic mass units (u)

  2. based on the mass of 1 atom of carbon-12

  3. different for different isotopes of an element

  4. all of the above are true


Correct Option: D
Explanation:

Relative atomic mass is the mass of an atom measured relative to 1/12th the mass of 1 atom of C-12 isotope which is also known as atomic mass unit or amu(u).

Chromium has four stable isotopes.$4.31$% Cr-$50$, $83.76$% Cr-$52$, $9.55$% Cr-$53$ and $2.38$% Cr-$54$. Calculate the relative atomic mass?

  1. $52.1$

  2. $52.4$

  3. $51.8$

  4. None


Correct Option: A
Explanation:

Average atomic mass is given by:


$M _{av}=\dfrac{50\times0.0431+52\times0.8376+53\times0.0955+54\times0.0238}{1}=52.1\ gm/mol$

Consider a reversible isentropic expansion of $1$ mole of an ideal monoatomic gas from ${27}^{o}C$ to ${927}^{o}C$. If the initial pressure of gas was $1$ bar, the final pressure of gas becomes

  1. $4$ bar

  2. $8$ bar

  3. $0.125$ bar

  4. $0.25$ bar


Correct Option: B

Boron found in nature has an atomic weight of 10.811 and is made up of the isotopes $\displaystyle { B }^{ 10 }$ (mass 10.013 amu) and $\displaystyle { B }^{ 11 }$ (mass 11.0093). What percentage of naturally occurring boron is made up of $\displaystyle { B }^{ 10 }$ and $\displaystyle { B }^{ 11 }$, respectively?

  1. 30 : 70

  2. 25 : 75

  3. 20 : 80

  4. 15 : 85


Correct Option: C
Explanation:

Let the abundance of $B^{11}$ be $x$% and $B^{10}$ be $(100-x)$%

Average atomic mass= [Atomic mass of $B^{11} \times$ abundance + Atomic mass of $B^{10}\times$ abundance]$/100$ 
$\Rightarrow 10.811=\cfrac { 11.0093\times x(percent)+10.013\times (100-x)(percent) }{ 100 } $
$\Rightarrow 10.811\times 100= 11.0093x$%$+1001.3-10.013x$%
$\Rightarrow 1081.1=0.9963x$%$+1001.3$
$\Rightarrow 1081.1-1001.3=0.9963x$%
$\Rightarrow 79.8=0.9963x$%
$\Rightarrow x$%=$\cfrac {79.8}{0.9963}$
$\therefore x$%=$80$
$(100-x)$%=$20$
$\therefore$ Natural abundance of $B^{10}=20$
    Natural abundance of $B^{11}=80$
Ratio= $20:80$

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