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The threshold odour number (TON) for a water sample of 40 ml, dilute to standard 200 mlmixture, in which odour is just barely detectable to the sense of smell, is
- a)8
- b)5
- c)50
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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To determine the threshold odor number (TON) for a water sample, we need to dilute it to a standard mixture and determine the point at which the odor is just barely detectable to the sense of smell. Let's break down the steps to find the correct answer:
1. Understanding the Threshold Odor Number (TON):
The TON is a measure of the ability of water to produce an odor. It represents the dilution factor required to make the odor just barely detectable. A lower TON indicates a stronger odor.
2. Given Data:
- Volume of water sample = 40 ml
- Volume of standard mixture = 200 ml
3. Calculation:
To find the TON, we need to determine the dilution factor. Since the water sample is diluted to a standard mixture, the dilution factor is given by the ratio of the volume of the standard mixture to the volume of the water sample:
Dilution factor = Volume of standard mixture / Volume of water sample
= 200 ml / 40 ml
= 5
The dilution factor of 5 means that the water sample is diluted 5 times to make the standard mixture. Therefore, the correct answer is option 'B' - 5.
4. Explanation:
Based on the dilution factor, a TON of 5 means that the odor of the water sample is just barely detectable when it is diluted 5 times. In other words, the water sample has a relatively strong odor, requiring less dilution to reach the threshold level.
It's important to note that the TON is a subjective measure and can vary depending on the sensitivity of the individuals involved in the odor testing. However, in this particular scenario, the correct answer is option 'B' - 5 based on the given dilution factor.
1. Understanding the Threshold Odor Number (TON):
The TON is a measure of the ability of water to produce an odor. It represents the dilution factor required to make the odor just barely detectable. A lower TON indicates a stronger odor.
2. Given Data:
- Volume of water sample = 40 ml
- Volume of standard mixture = 200 ml
3. Calculation:
To find the TON, we need to determine the dilution factor. Since the water sample is diluted to a standard mixture, the dilution factor is given by the ratio of the volume of the standard mixture to the volume of the water sample:
Dilution factor = Volume of standard mixture / Volume of water sample
= 200 ml / 40 ml
= 5
The dilution factor of 5 means that the water sample is diluted 5 times to make the standard mixture. Therefore, the correct answer is option 'B' - 5.
4. Explanation:
Based on the dilution factor, a TON of 5 means that the odor of the water sample is just barely detectable when it is diluted 5 times. In other words, the water sample has a relatively strong odor, requiring less dilution to reach the threshold level.
It's important to note that the TON is a subjective measure and can vary depending on the sensitivity of the individuals involved in the odor testing. However, in this particular scenario, the correct answer is option 'B' - 5 based on the given dilution factor.
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The threshold odour number (TON) for a water sample of 40 ml, dilute t...
Threshold odour number represent the dilution ratio at which the odour is hardly detectable.
T.O.N. = 200/40 = 5
T.O.N. = 200/40 = 5
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The threshold odour number (TON) for a water sample of 40 ml, dilute to standard 200 mlmixture, in which odour is just barely detectable to the sense of smell, isa)8b)5c)50d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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