Tag: determination of atomic and isotopic mass

Questions Related to determination of atomic and isotopic mass

 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= $20:80$