A water sample has ppm level concentration of following anions
`F^(-)=10 , SO_(4)^(2-)=100 , NO_(3)^(-) =50`
the anion/anions that make/makes the water sample unsuitable for drinking is/are

Options

A. `10:1`
B. `4:5`
C. `5:4`
D. `1:10`

Answer

To determine which anion(s) in the given water sample make it unsuitable for drinking, we will analyze the concentration of each anion against its permissible limit.

### Step-by-Step Solution:

1. **Identify the Anions and Their Concentrations:**
- Fluoride (F⁻) = 10 ppm
- Sulfate (SO₄²⁻) = 100 ppm
- Nitrate (NO₃⁻) = 50 ppm

2. **Know the Permissible Limits for Drinking Water:**
- Permissible limit for Fluoride (F⁻) = 1.5 ppm
- Permissible limit for Sulfate (SO₄²⁻) = 500 ppm
- Permissible limit for Nitrate (NO₃⁻) = 50 ppm

3. **Compare Each Anion's Concentration with Its Limit:**
- For Fluoride (F⁻):
- Given: 10 ppm
- Limit: 1.5 ppm
- Conclusion: 10 ppm > 1.5 ppm (Exceeds limit)

- For Sulfate (SO₄²⁻):
- Given: 100 ppm
- Limit: 500 ppm
- Conclusion: 100 ppm < 500 ppm (Within limit)

- For Nitrate (NO₃⁻):
- Given: 50 ppm
- Limit: 50 ppm
- Conclusion: 50 ppm = 50 ppm (At the limit, but acceptable)

4. **Determine Unsuitable Anions:**
- From the comparisons:
- Fluoride (F⁻) is unsuitable as it exceeds the permissible limit.
- Sulfate (SO₄²⁻) and Nitrate (NO₃⁻) are within acceptable limits.

5. **Final Conclusion:**
- The anion that makes the water sample unsuitable for drinking is **Fluoride (F⁻)**.

### Answer:
The anion that makes the water sample unsuitable for drinking is **Fluoride (F⁻)**.

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