The density of a salt solution is 1.13g`cm^(−3)` and it contains 20% of NaCI by weight The volume of the solution containing 46.0 g of the salt will be

Explanation

To solve the problem, we need to find the volume of a salt solution that contains 46.0 g of NaCl, given that the density of the solution is 1.13 g/cm³ and it contains 20% NaCl by weight.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Density of the solution = 1.13 g/cm³
- Percentage of NaCl by weight = 20%
- Mass of NaCl = 46.0 g

2. **Set Up the Equation:**
- Let the volume of the solution be \( V \) cm³.
- The mass of the solution can be calculated using the density formula:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.13 \times V \text{ grams} \]

3. **Relate the Mass of NaCl to the Mass of the Solution:**
- Since the solution is 20% NaCl by weight, we can express this as:
\[ \text{Mass of NaCl} = 20\% \text{ of Mass of solution} \]
- This can be written mathematically as:
\[ \frac{20}{100} \times \text{Mass of solution} = 46.0 \text{ g} \]
- Substituting the mass of the solution from step 2:
\[ \frac{20}{100} \times (1.13 \times V) = 46.0 \]

4. **Solve for Volume \( V \):**
- Rearranging the equation gives:
\[ 0.20 \times 1.13 \times V = 46.0 \]
- Simplifying further:
\[ 0.226 \times V = 46.0 \]
- Now, divide both sides by 0.226 to find \( V \):
\[ V = \frac{46.0}{0.226} \approx 203.5 \text{ cm}^3 \]

5. **Conclusion:**
- The volume of the solution containing 46.0 g of NaCl is approximately **203.5 cm³**.