Question

An ion with mass number 37 contains 1unit negative charge, if the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Answer 1

See bro
Let the no. Of electrons be x
No. Of neutron = x + 11.1% of x
Now no. Of protons = x-1
Niw as we know that
A=p+n
37=x+11.1%x + x-1
37= 2x + 111/1000x - 1
37000= 2000x + 111x - 1000
38000=2111x
X=18
Therefore we know from above
P=x-1
Put value of x
P=17

So answer will be chlorine i.e chlorine anion

Answer 2

$\fbox\textcolor{orange}{\texttt{Answer = Cl(37,17)}}$

$\textcolor{indigo}{\texttt{Explanation :-}}$

Suppose no. of electrons = x

no. of neutrons = $x + \frac{11.1}{100}\times{x}\: = \:1.111x$

No. of electrons in the neutral atom = $x - 1$

No. of protons in the neutral atom =$x - 1$

Mass no. = no. of neutrons + no. of protons

ie, $37\:=\:1.111x+x-1$

$2.111x\:=\:37+1\:=\:38$

$x = \frac{38}{2.111}\:=\texttt{18}$

No. of protons = Atomic no. =

$x-1\:=\:18-1\:=\texttt{17}$

Hence the symbol of the ion is: $</h3><h3>\fbox{^{37}_{17}Cl}$