Question

If one mole of $H_{2}S{O}_4$ reacts with one mole of $NaOH$, equivalent weight of $H_{2}S{O}_4$ will be :

• A. $98$
• B. $49$
• C. $96$
• D. $48$

Answer 1

Answer: (A)

$98$

One mole of $H_{2}S{O}_4$ reacts with one mole of $NaOH$. It means that only one $H^{+}$ ion is reacting with the $O{H}^{-}$ ion present in $NaOH$.

$\therefore$ No. of replaceable $H^{+}$ ions= $1$

$\therefore$ Equivalent weight= $\frac{Molecularweight}{No.ofreplaceable{H}^{+}ions}=\frac{98}{1}=98$