Question

The isotopic abundance of ${}^{12}C\,and\,{}^{14}C\,is\,(98\% )\,and\,(2\% )$ respectively. What would be the number of ${}^{14}C$ isotope in $12\,grams\,$ of carbon sample?
(A) $1.03\, \times \,{10^{22}}$
(B) $3.01\,\, \times \,{10^{23}}$
(C) $5.88\,\, \times \,{10^{23}}$
(D) $6.02\,\, \times \,{10^{23}}$

Answer 1

Hint : An isotope refers to a form of an element that consists of the equal number of protons but distinct numbers of neutrons. Example: ${}^{14}C$ is a naturally occurring isotope of carbon which is radioactive. It has eight neutrons and six protons in the nucleus. The other example is ${}^{12}C\,$ which is a stable isotope having six neutrons and six protons.

Complete Step By Step Answer:
We know that molecular mass of ${}^{14}C$ is $14\,g$ as it is written just above the symbol. Similarly, the molecular mass of ${}^{12}C\,$ is $12\,g$ . First step is to find out the mass of each isotope.
$Molar\,mass\,of\,{}^{14}C = \,14g$
$Molar\,mass\,of\,{}^{12}C = \,12g$
Second step is to use the information of the number of particles in one mole of any substance. Any substance contains $6.023\, \times \,{10^{23}}\,atoms$ this is called the Avogadro number. Number of atoms present in ${}^{14}C$ is equals to,
$Number\,of\,atoms\, = \,6.023\, \times \,{10^{23}}\,atoms$
Step three is to multiply this amount with $12g$ so as to get the number of atoms in that gram. Like for number of atoms in $12g$ of ${}^{14}C$ sample is equals to,
$Number\,of\,atoms\,in\,12g\,of\,{}^{14}C = \,\dfrac{{12 \times 6.023\, \times \,{{10}^{23}}}}{{14}}\,atoms$
= $5.16\, \times \,{10^{23}}\,atoms$
After solving we get the amount of atoms as $5.16\, \times \,{10^{23}}\,atoms$ . We are given isotopes abundance of ${}^{14}C$ is $2\%$ it means that in nature it exists as only $2\%$ . That means number of ${}^{14}C$ atoms in $12g$ carbon sample becomes $2\%$ of $5.16\, \times \,{10^{23}}\,atoms$ thus, let’s solve for $2\%$ ,
$Number\,of\,atoms = \,2\% \, \times 5.16\, \times \,{10^{23}}\,atoms\, = \,\dfrac{2}{{100}}\, \times \,5.16\, \times \,{10^{23}}\,atoms$
$\, = 10.32 \times \,{10^{23}}\,atoms$
The value can be written according to the significant place that is decimal is after one digit.
$= 1.032 \times \,{10^{22}}\,atoms$
So, these are the number of ${}^{14}C$ isotope in $12\,grams\,$ of carbon sample.
Hence, correct option is A. i.e $\,1.032 \times \,{10^{22}}\,$

Note :
Carbon has a total of $15\,\,isotopes$ ranging from ${}^8C\,to\,{}^{22}C$ out of which ${}^{12}C\,$ and ${}^{13}C\,$ are found to be stable. ${}^{14}C\,$ is the only radioisotope of carbon that occurs in nature. We can calculate the weighted average mass (average atomic mass) of an element with the information of their isotopic abundance. For instance, if we want to calculate the weighted average mass of a carbon element in the present case, then we have to multiply the molecular mass of carbon with its relative abundance.