\(\Rightarrow 23 \times 2+12+48+18 x\)
\(\Rightarrow 46+12+48+18 x\)
\(\Rightarrow(106+18 x )\)
\(Eqwt =\frac{ M }{2}=(53+9 x )\)
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 \(\quad 10^{-2}=\frac{1.43}{53+9 x }\)
\(53+9 x =143\)
\(9 x =90\)
\(x =10.00\)
(આણ્વિય દળ $\left.\mathrm{H}_{2} \mathrm{SO}_{4}=98 \;\mathrm{g} / \mathrm{mol}\right)$