Solve the following


In basic medium $\mathrm{CrO}_{4}^{2-}$ oxidizes $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ to form $\mathrm{SO}_{4}^{2-}$ and itself changes into $\mathrm{Cr}(\mathrm{OH})_{4}^{-}$. The volume of $0.154 \mathrm{MCrO}_{4}^{2-}$ required to react with $40 \mathrm{~mL}$ of $0.25 \mathrm{MS}_{2} \mathrm{O}_{3}{ }^{2-}$ is _______________. (Rounded-off to the nearest integer)



$17 \mathrm{H}_{2} \mathrm{O}+8 \mathrm{CrO}_{4}+3 \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 6 \mathrm{SO}_{4}+8 \mathrm{Cr}(\mathrm{OH})_{4}^{-}+2 \mathrm{OH}^{-}$

Applying mole-mole analysis $\frac{0.154 \times \mathrm{V}}{8}=\frac{40 \times 0.25}{3}$

$V=173 \mathrm{~mL}$

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