How many moles of  ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$  can be reduced by 1 mole of 

 ${\mathrm{Sn}}^{2+}$ in acidic medium.

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6e^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

$\left({\mathrm{Sn}}^2\to {\mathrm{Sn}}^{4+}+2e^{-}\right) 3$

$\overline{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+3\stackrel{ }{{\mathrm{Sn}}^{2+}}\to 3{\mathrm{Sn}}^{4+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}}$

It is clear from this equation that 3 moles of ${\text{Sn}}^{2+}$ reduce one 

mole of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ hence 1 mol. of  ${\mathrm{Sn}}^{+2}$ will   reduce  $\frac{1}{3}\text{ moles of }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$