Home
Class 11
CHEMISTRY
500 ml of sample COD of water required 1...

500 ml of sample COD of water required 19.6 mg of `K_(2)Cr_(2)O_(7)` of water sample is

A

8 ppm

B

6.4 ppm

C

16.8 ppm

D

4.9 ppm

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the Chemical Oxygen Demand (COD) of the water sample, we can follow these steps: ### Step 1: Understand the Given Information - Volume of water sample = 500 ml - Mass of K₂Cr₂O₇ used = 19.6 mg - Molecular weight of K₂Cr₂O₇ = 294 g/mol ### Step 2: Convert the Mass of K₂Cr₂O₇ to Grams Since the mass is given in milligrams, we need to convert it to grams: \[ 19.6 \, \text{mg} = \frac{19.6}{1000} \, \text{g} = 0.0196 \, \text{g} \] ### Step 3: Calculate Moles of K₂Cr₂O₇ Using the molecular weight, we can find the number of moles of K₂Cr₂O₇: \[ \text{Moles of K₂Cr₂O₇} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.0196 \, \text{g}}{294 \, \text{g/mol}} \] Calculating this gives: \[ \text{Moles of K₂Cr₂O₇} = 6.67 \times 10^{-5} \, \text{mol} \] ### Step 4: Determine Moles of Nascent Oxygen Produced From the balanced reaction, we know that 1 mole of K₂Cr₂O₇ produces 3 moles of nascent oxygen. Therefore, the moles of nascent oxygen produced are: \[ \text{Moles of nascent oxygen} = 3 \times \text{Moles of K₂Cr₂O₇} = 3 \times 6.67 \times 10^{-5} \, \text{mol} \] Calculating this gives: \[ \text{Moles of nascent oxygen} = 2.00 \times 10^{-4} \, \text{mol} \] ### Step 5: Convert Moles of Nascent Oxygen to Grams The molar mass of oxygen (O) is approximately 16 g/mol. Therefore, the mass of nascent oxygen produced is: \[ \text{Mass of nascent oxygen} = \text{Moles} \times \text{Molar mass} = 2.00 \times 10^{-4} \, \text{mol} \times 16 \, \text{g/mol} \] Calculating this gives: \[ \text{Mass of nascent oxygen} = 0.0032 \, \text{g} = 3.2 \, \text{mg} \] ### Step 6: Calculate COD for 1000 ml Since the 3.2 mg of nascent oxygen corresponds to the 500 ml sample, we can find the COD for 1000 ml: \[ \text{COD for 1000 ml} = \frac{3.2 \, \text{mg}}{500 \, \text{ml}} \times 1000 \, \text{ml} = 6.4 \, \text{mg} \] ### Conclusion The Chemical Oxygen Demand (COD) of the water sample is **6.4 mg**.

To calculate the Chemical Oxygen Demand (COD) of the water sample, we can follow these steps: ### Step 1: Understand the Given Information - Volume of water sample = 500 ml - Mass of K₂Cr₂O₇ used = 19.6 mg - Molecular weight of K₂Cr₂O₇ = 294 g/mol ### Step 2: Convert the Mass of K₂Cr₂O₇ to Grams ...
Promotional Banner

Topper's Solved these Questions

  • ENVIRONMENTAL CHEMISTRY

    NARAYNA|Exercise Excercise - II (a) (C.W) (Air Pollution)|14 Videos
  • ENVIRONMENTAL CHEMISTRY

    NARAYNA|Exercise EXCERCISE - II (a) (H.W) ( Environmental segments, Definitions & Terms)|3 Videos
  • ENVIRONMENTAL CHEMISTRY

    NARAYNA|Exercise EXCERCISE- I (H.W) (Water pollution)|21 Videos
  • CLASSIFICATION OF ELEMENTS AND PERIODICITY

    NARAYNA|Exercise EXERCISE - 4|17 Videos
  • F-BLOCK ELEMENTS

    NARAYNA|Exercise LEVEL-II (C.W)|1 Videos

Similar Questions

Explore conceptually related problems

500 ml of a sample COD of water required 19.6 mg of K_(2)Cr_(2)O_(7) of water sample is

100 mL of a sample of hard water requires 25.1 mL of 0.02 N H_(2)SO_(4) for complete reaction, The hardness of water ( density 1g//mL) is:

100 ml of a sample of water requires 1.96 mg of potassium dichromate in the presence of 50% H_(2)SO_(4) for the oxidation of dissolved oxygen demand.

Potassium iodide reacts with acidified K_(2)Cr_(2)O_(7) . How many moles of KI are required for one mole of K_(2)Cr_(2)O_(7) ?

A 200 mL sample of hard water requires 33.0 " mL of " 0.01 M H_2SO_4 for complete neutralisation. 200 " mL of " the same sample was boiled with 15.0 " mL of " 0.1 M NaOH solution, filtered and made up to 200 mL again. This sample now requires 53.6 " mL of " 0.01 M H_2SO_4 . Calculate Mg hardness.