The equivalent weight of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the following reaction is $\mathrm{Na}_{2}…

The equivalent weight of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the following reaction is
$\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+3 \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}ight)_{3}+\mathrm{H}_{2} \mathrm{O}$
  1. $98$
  2. $\frac{98}{6}$
  3. $\frac{98}{2}$
  4. $\frac{98}{8}$

Solution

In this reaction, $6 \mathrm{~mol}$ of $e^{-}$ are involved with
$1 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in this redox reaction. $\mathrm{H}_{2} \mathrm{SO}_{4}$ acts here as acidic medium $\left({ }^{\prime} n^{\prime}=2ight),\left(2 \mathrm{H}^{\oplus}ight)$.
So, $E w$ of $\mathrm{H}_{2} \mathrm{SO}_{4}=\frac{M}{2}=\frac{98}{2}=49$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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