Tag: determination of atomic and isotopic mass

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Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

 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. $186.1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Relative atomic mass is calculated as (abundance1 * mass1 + abundance2 * mass2) / 100. For Rhenium: (37.1 * 185 + 62.9 * 187) / 100 = (6863.5 + 11762.3) / 100 = 18625.8 / 100 = 186.258, which rounds to 186.3.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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).

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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= $20:80$