The concentration of iron, lead ,nitrate, and fluoride in a water sample from an underground lake was found to be 0.18 ppm, 35 ppb, 80 ppm and 1000 ppb, 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 makes the water unsuitable for drinking, we need to compare the concentrations of iron, lead, nitrate, and fluoride in the water sample. Here’s a step-by-step solution:

### Step 1: Identify the concentrations
We have the following concentrations:
- Iron: 0.18 ppm
- Lead: 35 ppb
- Nitrate: 80 ppm
- Fluoride: 1000 ppb

### Step 2: Convert all concentrations to the same unit
To compare these values effectively, we should convert all concentrations to parts per million (ppm).

1. **Convert Lead from ppb to ppm**:
\[ \text{Lead (ppm)} = \frac{35 \text{ ppb}}{1000} = 0.035 \text{ ppm} \]

2. **Convert Fluoride from ppb to ppm**:
\[ \text{Fluoride (ppm)} = \frac{1000 \text{ ppb}}{1000} = 1 \text{ ppm} \]

### Step 3: List the concentrations in ppm
Now we can list all the concentrations in ppm:
- Iron: 0.18 ppm
- Lead: 0.035 ppm
- Nitrate: 80 ppm
- Fluoride: 1 ppm

### Step 4: Compare the concentrations
Now we compare the concentrations:
- Iron: 0.18 ppm
- Lead: 0.035 ppm
- Nitrate: 80 ppm
- Fluoride: 1 ppm

The highest concentration is that of **nitrate**, which is **80 ppm**.

### Step 5: Conclusion
Since the concentration of nitrate (80 ppm) is significantly higher than the other substances, it is the reason why the water is unsuitable for drinking.

### Final Answer
The water is unsuitable for drinking due to the high concentration of **nitrate**.

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