
Average atomic weight of an element M is$51.7$. If two isotopes of M, ${M^{50}}$and ${M^{52}}$are present, then calculate the percentage of occurrence of ${M^{50}}$in nature.
Answer
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Hint: The average weight is defined as the sum of weights of all the isotopes multiplied by the percentage fraction of their presence in the atmosphere. Mathematically,
Average atomic weight = ${F_1}{M_1} + {F_2}{M_2} + {F_3}{M_3} + ..... + {F_n}{M_n}$
Where ${F_1},{\text{ }}{F_2},{\text{ }}{F_3},{\text{ }}{F_n}$ correspond to percentage fraction
${M_1},{\text{ }}{{\text{M}}_2},{\text{ }}{{\text{M}}_3},{\text{ }}{{\text{M}}_n}$correspond to molecular weights of isotopes
We have the average weight, thus we can find the percentage of one isotope.
Complete step by step solution:
First, let us see what isotopes are.
The isotopes are the atoms of the same element that have the same atomic number but different mass number. In nature the sum of total conc. of all the isotopes of an element is considered $100\% $.
The average weight is the total sum of weights of all the isotopes multiplied by the percentage fraction of their presence. Mathematically, it can be written as -
Average atomic weight = ${F_1}{M_1} + {F_2}{M_2} + {F_3}{M_3} + ..... + {F_n}{M_n}$
Where ${F_1},{\text{ }}{F_2},{\text{ }}{F_3},{\text{ }}{F_n}$ correspond to percentage fraction
${M_1},{\text{ }}{{\text{M}}_2},{\text{ }}{{\text{M}}_3},{\text{ }}{{\text{M}}_n}$correspond to molecular weights of isotopes
We have average weight of M = 51.7
The two isotopes are ${M^{50}}$and ${M^{52}}$
Thus, we can write -
$51.7 = x \times 50 + (1 - x) \times 52$
$51.7 = 50x + 52 - 52x$
$ - 2x = - 0.3$
$x = \frac{{0.3}}{2}$
$x = 0.15$
The value of x corresponds to the amount of ${M^{50}}$. So, the ${M^{50}}$is $0.15$ in nature i.e. $15\% $.
So, the abundance of ${M^{50}}$is $15\% $.
Note: The weight that we generally write of the element is the average weight only which is calculated by addition of weights of all isotopes multiplied by their relative concentration w.r.t each other in the atmosphere.
Average atomic weight = ${F_1}{M_1} + {F_2}{M_2} + {F_3}{M_3} + ..... + {F_n}{M_n}$
Where ${F_1},{\text{ }}{F_2},{\text{ }}{F_3},{\text{ }}{F_n}$ correspond to percentage fraction
${M_1},{\text{ }}{{\text{M}}_2},{\text{ }}{{\text{M}}_3},{\text{ }}{{\text{M}}_n}$correspond to molecular weights of isotopes
We have the average weight, thus we can find the percentage of one isotope.
Complete step by step solution:
First, let us see what isotopes are.
The isotopes are the atoms of the same element that have the same atomic number but different mass number. In nature the sum of total conc. of all the isotopes of an element is considered $100\% $.
The average weight is the total sum of weights of all the isotopes multiplied by the percentage fraction of their presence. Mathematically, it can be written as -
Average atomic weight = ${F_1}{M_1} + {F_2}{M_2} + {F_3}{M_3} + ..... + {F_n}{M_n}$
Where ${F_1},{\text{ }}{F_2},{\text{ }}{F_3},{\text{ }}{F_n}$ correspond to percentage fraction
${M_1},{\text{ }}{{\text{M}}_2},{\text{ }}{{\text{M}}_3},{\text{ }}{{\text{M}}_n}$correspond to molecular weights of isotopes
We have average weight of M = 51.7
The two isotopes are ${M^{50}}$and ${M^{52}}$
Thus, we can write -
$51.7 = x \times 50 + (1 - x) \times 52$
$51.7 = 50x + 52 - 52x$
$ - 2x = - 0.3$
$x = \frac{{0.3}}{2}$
$x = 0.15$
The value of x corresponds to the amount of ${M^{50}}$. So, the ${M^{50}}$is $0.15$ in nature i.e. $15\% $.
So, the abundance of ${M^{50}}$is $15\% $.
Note: The weight that we generally write of the element is the average weight only which is calculated by addition of weights of all isotopes multiplied by their relative concentration w.r.t each other in the atmosphere.
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