In basic medium $\mathrm{CrO}_{4}^{2-}$ oxidises $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ to form $\mathrm{SO}_{4}{ }^{2-}$ and itself changes into $\mathrm{Cr}(\mathrm{OH})_{4}^{-}$. The volume of $0.154 \mathrm{M} \mathrm{CrO}_{4}^{2-}$ required to react with $40 \mathrm{~mL}$ of $0.25 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is_____________ $\mathrm{mL}$. (Rounded-off to the nearest integer)
$\stackrel{+6}{\mathrm{CrO}_{4}^{2-}}+\stackrel{+2}{\mathrm{~S}}_{2} \mathrm{O}_{3}^{2-} \rightarrow \stackrel{+6}{\mathrm{~S} \mathrm{O}_{4}^{2-}}+\stackrel{+3}{\mathrm{Cr}}(\mathrm{OH})_{4}^{-}$
gm equi. of $\mathrm{CrO}_{4}^{2-}=\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$
$0.154 \times 3 \times v=0.25 \times 40 \times 8$
$\mathrm{v}=173.16=173 \mathrm{ml}$
Hence answer is (173)