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JEE Advanced levelNumerical answerDifficulty 3 of 3ev154

In a COD determination, $25.0 mL$ of a sample is refluxed with $20.0 mL$ of $0.250 N$ $K X_{2} Cr X_{2} O X_{7}$ in acid. The unused dichromate requires $14.6 mL$ of $0.100 N$ ferrous ammonium sulfate in the back titration. Use 1 equivalent of $O X_{2}$ = 8 g. Give the COD in $mg O_{2}$ per litre of sample to the nearest integer.

Numerical answer type, answer in mg/L, tolerance 0.5.

Answer: 1133 mg/L

Dichromate added =

20.0 \times 0.250 = 5.00

meq. Excess found by back titration =

14.6 \times 0.100 = 1.46

meq, so dichromate used =

3.54

meq. This is equal to

3.54

meq of

O X_{2}

, i.e.

3.54 \times 8 = 28.32

mg per 25.0 mL. COD

= 28.32 \times 1000/25.0 = 1132.8 \approx 1133

mg/L.

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