In basic medium $Cr{O_4}^{2 - }$ oxidises ${S_2}{O_3}^{2 - }$ to form $S{O_4}^{2 - }$ and itself changes into $Cr{(OH)_4}^ - $. The volume of 0.154 M $Cr{O_4}^{2 - }$ required to react with 40 mL of 0.25 M ${S_2}{O_3}^{2 - }$ is __________ mL. (Rounded off to the nearest integer)