In chlorine gas, the ratio of $C{{l}^{35}}$and $C{{l}^{37}}$is?
A. $1:3$
B. $3:1$
C. $1:1$
D. $1:4$
Answer
623.4k+ views
Hint: As per the given question, chlorine is having two isotopes that are $C{{l}^{35}}$and $C{{l}^{37}}$. But, the relative atomic mass of chlorine is not 36 instead it is $35.5$. So, by using the direct formula of relative atomic mass, we can find out what percentage of the atoms i.e. $C{{l}^{35}}$and $C{{l}^{37}}$are present and further we will get the ratio.
Complete step by step solution:
Isotopes are referred to as the variants of chemical elements which possess the same number of protons and electrons but will have a different number of neutrons. In short, they will have the same atomic number but different mass number. While considering their atomic mass, we have to take into account their relative atomic mass.
So, as per the given question, the chlorine gas has $C{{l}^{35}}$and $C{{l}^{37}}$in an unknown ratio. Thus, we have to find out the ratio of two isotopes.
Applying the formula of relative atomic mass of chlorine, we can get the ratio of two isotopes.
So, the formula for relative atomic mass is,
\[Average Atomic Mass=Atomic Weight\times Percentage Abundance\]
And atomic mass of chlorine will be,
\[Average Atomic Mass=\dfrac{Atomic Mass Of(C{{l}^{35}}+C{{l}^{37}})}{Number Of Moles(total)}\]
Let’s consider, the percentage abundance of $C{{l}^{37}}$be ‘X’ and
The percentage abundance of $C{{l}^{35}}$be 1.
We know that the relative or average atomic mass of chlorine is \[35.5\].
And, the atomic mass of $C{{l}^{37}}$is $37X$ while that of $C{{l}^{35}}$is 35.
Total number of moles will be $X+1$.
So, using the formula of average atomic mass of chlorine, we will get the value of X.
So, $35.5=\dfrac{37\times X+35\times 1}{X+1}$
Then, $35.5(X+1)=37X+35$
Then, $1.5X=0.5$
So,
$X=\dfrac{1}{3}$
Thus, the ratio of $C{{l}^{35}}$and $C{{l}^{37}}$will be,
$\dfrac{C{{l}^{35}}}{C{{l}^{37}}}=\dfrac{3}{1}$
Hence, the correct option is B.
Note: Relative atomic mass, denoted by ${{A}_{r}}$ is a dimensionless physical quantity which can be defined as the ratio of the average mass of atoms of an element in a given sample to the atomic mass constant (i.e. $\dfrac{1}{12}$of the mass of a carbon-12 atom). For isotopes, we have to consider the relative atomic mass as the atomic weight for that element.
Complete step by step solution:
Isotopes are referred to as the variants of chemical elements which possess the same number of protons and electrons but will have a different number of neutrons. In short, they will have the same atomic number but different mass number. While considering their atomic mass, we have to take into account their relative atomic mass.
So, as per the given question, the chlorine gas has $C{{l}^{35}}$and $C{{l}^{37}}$in an unknown ratio. Thus, we have to find out the ratio of two isotopes.
Applying the formula of relative atomic mass of chlorine, we can get the ratio of two isotopes.
So, the formula for relative atomic mass is,
\[Average Atomic Mass=Atomic Weight\times Percentage Abundance\]
And atomic mass of chlorine will be,
\[Average Atomic Mass=\dfrac{Atomic Mass Of(C{{l}^{35}}+C{{l}^{37}})}{Number Of Moles(total)}\]
Let’s consider, the percentage abundance of $C{{l}^{37}}$be ‘X’ and
The percentage abundance of $C{{l}^{35}}$be 1.
We know that the relative or average atomic mass of chlorine is \[35.5\].
And, the atomic mass of $C{{l}^{37}}$is $37X$ while that of $C{{l}^{35}}$is 35.
Total number of moles will be $X+1$.
So, using the formula of average atomic mass of chlorine, we will get the value of X.
So, $35.5=\dfrac{37\times X+35\times 1}{X+1}$
Then, $35.5(X+1)=37X+35$
Then, $1.5X=0.5$
So,
$X=\dfrac{1}{3}$
Thus, the ratio of $C{{l}^{35}}$and $C{{l}^{37}}$will be,
$\dfrac{C{{l}^{35}}}{C{{l}^{37}}}=\dfrac{3}{1}$
Hence, the correct option is B.
Note: Relative atomic mass, denoted by ${{A}_{r}}$ is a dimensionless physical quantity which can be defined as the ratio of the average mass of atoms of an element in a given sample to the atomic mass constant (i.e. $\dfrac{1}{12}$of the mass of a carbon-12 atom). For isotopes, we have to consider the relative atomic mass as the atomic weight for that element.
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