$\Rightarrow 23 \times 2+12+48+18 x$
$\Rightarrow 46+12+48+18 x$
$\Rightarrow(106+18 x )$
$Eqwt =\frac{ M }{2}=(53+9 x )$
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 $\quad 10^{-2}=\frac{1.43}{53+9 x }$
$53+9 x =143$
$9 x =90$
$x =10.00$
($k _{ f }=1.86\,K\,kg\,mol ^{-1}$ )