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Question: B has two isotopes \[{}^{{\text{10}}}{\text{B}}\left( {{\text{19% }}} \right){\text{, }}{}^{{\text{1...

B has two isotopes {}^{{\text{10}}}{\text{B}}\left( {{\text{19% }}} \right){\text{, }}{}^{{\text{11}}}{\text{B}}\left( {{\text{81% }}} \right). The atomic mass of B is
A. 10.8110.81
B. 105{10^5}
C. 1111
D. 10.510.5

Explanation

Solution

We have to know that in chemistry, periodic tables play a vital role. In the periodic table there are totally 118118 elements. In the periodic table there are totally 1818 columns and 77 rows. The columns are called groups. Hence, 1818 groups in the periodic table. The rows are called periods. Hence, totally 77 period in the table. Boron is one of the elements in the periodic table. The isotopes mean two atoms having the same atomic number but different mass number and also having the same atomic symbol. Here 10B{}^{{\text{10}}}{\text{B}} and 11B{}^{11}B having same atomic number of five and same symbol of B but having different mass number of 1010 and 1111 .
Formula used:
The average atomic mass of the atom is defined as the average of the atomic mass of the atom with respect to their percentage in nature.
Average atomic mass is equal to the sum of the product mass of atom respect that counts in percentage.
average atomic mass = atomic weight of the atom×number of atoms count100{\text{average atomic mass = }}\dfrac{{{\text{atomic weight of the atom}} \times {\text{number of atoms count}}}}{{{\text{100}}}}

Complete answer:
There are two isotopes in boron. There are 10B{}^{{\text{10}}}{\text{B}} and 11B{}^{11}B.
The percentage of 10B{}^{{\text{10}}}{\text{B}} is {\text{19% }}.
The percentage of 11B{}^{{\text{11}}}{\text{B}} is {\text{81% }}.
The average atomic mass of the given data of boron is,
average atomic mass = atomic weight of the atom×number of atoms count100{\text{average atomic mass = }}\dfrac{{{\text{atomic weight of the atom}} \times {\text{number of atoms count}}}}{{{\text{100}}}}
Now we can substitute the given values we get,
=(10×19)+(11×81)100= \dfrac{{(10 \times 19) + (11 \times 81)}}{{100}}
On simplification we get,
=10.81= 10.81
Hence, B has two isotopes {}^{{\text{10}}}{\text{B}}\left( {{\text{19% }}} \right) , {}^{{\text{11}}}{\text{B}}\left( {{\text{81% }}} \right) . The atomic mass of B is 10.8110.81.

Option A is the correct answer.

Note:
We have to remember that the atomic number of the element is nothing but the number of electrons or number of protons. The mass number of the atom is nothing but the sum of the number of protons and the number of neutrons. Boron is one of the elements in the periodic table. The atomic number of boron is 55. The symbol of boron is B{\text{B}}. The mass number of boron is 10.