The volume V of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ 7 (molar mass $294\mathrm{g}{\mathrm{mol}}^{-1}$) solution (mass concentration $0.005\mathrm{g}\text{ per }\mathrm{mL}$) is equivalent to $35.0\mathrm{mL}$ of  $0.02\mathrm{M}{\mathrm{KMnO}}_4$(molar mass  $158\mathrm{g}{\mathrm{mol}}^{-1}$), when these solutions are used in the titration in acid solution. The value of V would be

The reactions are  

$\frac{1}{6}{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+\frac{14}{6}{\mathrm{H}}^{+}+{\mathrm{e}}^{-}⟶\frac{1}{3}{\mathrm{Cr}}^{3+}+\frac{7}{6}{\mathrm{H}}_2\mathrm{O}$ and  $\frac{1}{5}{\mathrm{MnO}}_4^{-}+\frac{8}{5}{\mathrm{H}}^{+}+{\mathrm{e}}^{-}⟶\frac{1}{5}{\mathrm{Mn}}^{2+}+\frac{4}{5}{\mathrm{H}}_2\mathrm{O}$

Thus, $n\left(\frac{1}{6}{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right)\equiv n\left(\frac{1}{5}{\mathrm{MnO}}_4^{-}\right)\text{i.e. }6n\left({\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\right)=5n\left({\mathrm{MnO}}_4^{-}\right)$Now 

$n\left({\mathrm{KMnO}}_4\right)=VM=\left(0.035\mathrm{L}\right)\left(0.02{\mathrm{molL}}^{-1}\right)=0.0007\mathrm{mol}$

$n\left({\mathrm{K}}_2{\mathrm{CrO}}_7\right)=VM=V\left(\frac{0.005\mathrm{g}{\mathrm{mL}}^{-1}}{294\mathrm{g}{\mathrm{mol}}^{-1}}\right)=V\left(\frac{0.005}{294}\mathrm{mol}{\mathrm{mL}}^{-1}\right)$Thus $6V\left(\frac{0.005}{294}{\mathrm{molmLm}}^{-1}\right)=5\left(0.0007\mathrm{mol}\right)$  or  $V=\left(\frac{5}{6}\right)\left(\frac{0.0007\times 294}{0.005}\right)\mathrm{mL}=34.3\mathrm{mL}$