\(\Rightarrow 23 \times 2+12+48+18 x\)

\(\Rightarrow 46+12+48+18 x\)

\(\Rightarrow(106+18 x )\)

\(Eqwt =\frac{ M }{2}=(53+9 x )\)

As \(n _{\text {factor }}\) in dissolution will be determined from net cationic or anionic charge; which is \(2\) so

\(Eqwt =\frac{ M }{2}=53+9 x\)

Gmeq \(=\frac{ wt }{ Eqwt }=\frac{1.43}{53+9 x }\)

Normality \(=\frac{ Gmeq }{ V _{\text {litre }}}\)

Normality \(=0.1=\frac{1.43}{\frac{53+9 x }{0.1}}\)

As volume \(=100 ml\)

\(=0.1 Litre\)

So \(\quad 10^{-2}=\frac{1.43}{53+9 x }\)

\(53+9 x =143\)

\(9 x =90\)

\(x =10.00\)