Naturally occurring boron consists of two iso | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the percentage abundance of isotope $(I)$ be $x_1$ and isotope $(II)$ be $x_2$. Since $x_1 + x_2 = 100$,we can write $x_2 = 100 - x_1$.The ave

Vedclass · Accepted Answer

$20$ and $80$

Vedclass · Answer

(A) Let the percentage abundance of isotope $(I)$ be $x_1$ and isotope $(II)$ be $x_2$. Since $x_1 + x_2 = 100$,we can write $x_2 = 100 - x_1$.The average atomic mass is given by the formula: $	ext{Average Mass} = \frac{M_1 x_1 + M_2 x_2}{100}$.Substituting the given values: $10.81 = \frac{10.01 x_1 + 11.01 (100 - x_1)}{100}$.$1081 = 10.01 x_1 + 1101 - 11.01 x_1$.$1081 - 1101 = -1.00 x_1$.$-20 = -1.00 x_1$.$x_1 = 20$.Therefore,$x_2 = 100 - 20 = 80$.The percentage abundances are $20\%$ and $80\%$ respectively.

Naturally occurring boron consists of two isotopes with atomic masses $10.01 \ (I)$ and $11.01 \ (II)$. The average atomic mass of natural boron is $10.81$. Find the percentage abundance of isotopes $(I)$ and $(II)$ respectively.

Similar Questions

The specific heat of a metal is $0.16$,its approximate atomic weight would be

The unit of atomic mass,$amu$ is replaced by $u$,here $u$ stands for

The equivalent weight of an acid is equal to

What is meant by $1 \ amu$ or $1 \ u$?

In the transformation $FeC_2O_4 \rightarrow Fe^{3+} + CO_2$,the equivalent weight of $FeC_2O_4$ is $M/n$. The value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module