10 mL of water requires 1.47 mg of K2Cr2O 7 (M. wt. = 294) for oxidation of dissolved organic matter. C.O.D is

Question

10 mL of water requires 1.47 mg of K2Cr2O 7 (M. wt. = 294) for oxidation of dissolved organic matter. C.O.D is (A) 2.44 ppm (B) 24 ppm (C) 32 ppm (D) 1.6 ppm

Tardigrade Admin · Accepted Answer

10 mL water =1.47 × 10-3 g of K 2 Cr 2 O 7 106 mL water =1.47 g of K 2 Cr 2 O 7 ∵ 49 g of K 2 Cr 2 O 7 requires O 2=89 ∴ 147 g of K 2 Cr 2 O 7 will required O 2=(8 × 147/49) =24 g .

Q. $10 m L$ of water requires $1.47 m g$ of $K_{2} C r_{2} O_{7}$ (M. wt. $= 294$ ) for oxidation of dissolved organic matter. $C . O . D$ is

$2.44 pp m$

$24 pp m$

$32 pp m$

$1.6 pp m$

$10 m L$ water $= 1.47 \times 1 0^{- 3} g$ of $K_{2} C r_{2} O_{7}$
$1 0^{6} m L$ water $= 1.47 g$ of $K_{2} C r_{2} O_{7}$
$∵ 49 g$ of $K_{2} C r_{2} O_{7}$ requires $O_{2} = 89$
$∴ 147 g$ of $K_{2} C r_{2} O_{7}$ will required
$O_{2} = \frac{8 \times 147}{49}$
$= 24 g .$

Q. 10mL of water requires 1.47mg of K2​Cr2​O7​ (M. wt. =294) for oxidation of dissolved organic matter. C.O.D is

2.44ppm

24ppm

32ppm

1.6ppm