The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb,40 ppb,100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of

Options

A. Lead
B. Nitrate
C. Iron
D. Fluoride

Answer

To determine which substance in the water sample makes it unsuitable for drinking, we need to analyze the concentrations of fluoride, lead, nitrate, and iron provided in the question.

### Step-by-Step Solution:

1. **Identify the Concentrations**:
- Fluoride: 1000 ppb
- Lead: 40 ppb
- Nitrate: 100 ppm
- Iron: 0.2 ppm

2. **Convert ppb to ppm**:
- To compare the concentrations, we need to convert all values to the same unit. We know that:
- 1 ppm = 1000 ppb
- Therefore:
- Fluoride: \(1000 \, \text{ppb} = \frac{1000}{1000} \, \text{ppm} = 1 \, \text{ppm}\)
- Lead: \(40 \, \text{ppb} = \frac{40}{1000} \, \text{ppm} = 0.04 \, \text{ppm}\)

3. **List the Concentrations in ppm**:
- Fluoride: 1 ppm
- Lead: 0.04 ppm
- Nitrate: 100 ppm
- Iron: 0.2 ppm

4. **Identify the Highest Concentration**:
- Comparing the concentrations:
- Fluoride: 1 ppm
- Lead: 0.04 ppm
- Nitrate: 100 ppm
- Iron: 0.2 ppm
- The highest concentration is for **Nitrate** at **100 ppm**.

5. **Determine Unsuitability for Drinking**:
- According to health guidelines, high concentrations of certain substances can render water unsuitable for drinking. In this case, the high concentration of nitrate (100 ppm) makes the water sample unsuitable for drinking.

### Conclusion:
The water sample is unsuitable for drinking due to the high concentration of **nitrate**.