Test Index

NCERT Class XI Chemistry Structure of Atom Solutions

© examsnet.com
Question : 43 of 67
Marks: +1, -0
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.
Solution:  
Let number of electrons = x
According to the question, number of neutrons (n)
= x + x×11.1100\frac{x \times 11.1}{100} = 100x+11.1x100\frac{100x+11.1x}{100} = 111.1x100\frac{111.1x}{100} = 1.11x
We know that A = p + n
But p = x – 1
37 = x – 1 + 1.11x or 38 = 2.11x
∴ x = 382.11\frac{38}{2.11} = 18
So, number of electrons = 18.
Hence, number of protons = 18 – 1 = 17.
Thus, symbol of ion =  1737Cl−\,\mathrm{^{37}_{17}Cl^{-}}
© examsnet.com
Go to Question:
Prev Question
Next Question