In basic solution, $CrO_4^{2–}$ oxidises $S_2O_3^{2–}$ to form $Cr(OH)_4^{–} \& SO_4^{2–}$. The volume of $0.15\; M CrO_4^{2–}$ required of react with $40\; ml$ of $0.225\;M \;S_2O_3^{2–}$ would be :

Question

$\begin{array}{1 1} 20\;ml \\ 40\;ml \\ 80\;ml \\ 160\;ml \end{array} $