Calculate the COD of the effluent sample sample when 2 cm3 of the effluent requires 10.5 cm3 of 0.005 M K2Cr2O7 for complete oxidation.

According to the given question,

$1000\ cm^3 of\ 1M\ K_2Cr_2O_7 = 48\ gm\ of\ O_2$

$10.5\ cm^3 of\ 0.005M K_2Cr_2O_7 = \frac{48 \times 10.5 \times 0.005}{1000} = 0.00252\ gm\ of\ O_2$

$\therefore\ 2\ cm^3 of\ effluent\ contains = 0.00252\ gm\ of\ O_2 $

$So,\ 1000\ cm^3 of\ effluent\ contains = \frac{0.00252 \times 1000}{2} = 1.26\ gm\ of\ O_2$

$\therefore\ COD\ of\ effluent = 1.26 \times 1000 = 1260\ ppm\ Ans.$