40% (w/v) of NaCl solution (specific gravity = 1.12) is equivalent to

Options

A. `3.57xx10^5` ppm
B. `3.57xx10^6` ppm
C. `1xx10^6` ppm
D. `4xx10^5` ppm

Answer

To solve the problem of converting a 40% (w/v) NaCl solution with a specific gravity of 1.12 into parts per million (ppm), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding 40% (w/v)**:
- A 40% (w/v) solution means there are 40 grams of NaCl in 100 mL of the solution.

2. **Finding the Density**:
- The specific gravity of the solution is given as 1.12. Since specific gravity is the ratio of the density of a substance to the density of water (1 g/mL), we can say:
\[ \text{Density of solution} = \text{Specific Gravity} \times \text{Density of Water} = 1.12 \times 1 \text{ g/mL} = 1.12 \text{ g/mL} \]

3. **Calculating the Mass of the Solution**:
- For 100 mL of the solution, the mass can be calculated as:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.12 \text{ g/mL} \times 100 \text{ mL} = 112 \text{ g} \]

4. **Finding the Amount of NaCl in Larger Volume**:
- Now, we want to find out how much NaCl would be present in 1,000,000 grams (1 million grams) of the solution.
- We know that in 112 g of solution, there are 40 g of NaCl. We set up a proportion:
\[ \frac{40 \text{ g NaCl}}{112 \text{ g solution}} = \frac{x \text{ g NaCl}}{10^6 \text{ g solution}} \]
- Cross-multiplying gives:
\[ x = \frac{40 \times 10^6}{112} \]

5. **Calculating x**:
- Now we calculate \( x \):
\[ x = \frac{40 \times 10^6}{112} \approx 357142.857 \text{ g NaCl} \]

6. **Converting to ppm**:
- Since ppm (parts per million) is defined as grams of solute per million grams of solution, we can say:
\[ \text{ppm} = \frac{357142.857 \text{ g NaCl}}{10^6 \text{ g solution}} \approx 357.14 \text{ ppm} \]

7. **Final Result**:
- Therefore, the 40% (w/v) NaCl solution is equivalent to approximately \( 3.57 \times 10^5 \) ppm.