The density of a salt solution is 1.13g`cm^(−3)` and it contains 18% of NaCI by weight The volume of the solution containing 18.0 g of the salt will be

Options

A. 44`cm^3`
B. 88`cm^3`
C. 22`cm^3`
D. None

Explanation

To solve the problem, we need to find the volume of the salt solution that contains 18.0 g of NaCl, given that the density of the solution is 1.13 g/cm³ and it contains 18% NaCl by weight.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Density of the salt solution = 1.13 g/cm³
- Percentage of NaCl in the solution = 18% by weight
- Mass of NaCl = 18.0 g

2. **Calculate the Mass of the Solution:**
- Since the solution contains 18% NaCl by weight, we can set up the equation:
\[ \text{Mass of NaCl} = \text{Percentage of NaCl} \times \text{Mass of Solution} \]
- Let the mass of the solution be \( m \) grams. Then:
\[ 18\% \text{ of } m = 18.0 \text{ g} \]
- This can be expressed as:
\[ \frac{18}{100} \times m = 18.0 \]
- Rearranging gives:
\[ m = \frac{18.0 \times 100}{18} = 100 \text{ g} \]

3. **Calculate the Volume of the Solution:**
- Now that we have the mass of the solution, we can find the volume using the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
- Rearranging gives:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.13 \text{ g/cm}^3} \]
- Performing the calculation:
\[ \text{Volume} = \frac{100}{1.13} \approx 88.496 \text{ cm}^3 \]
- Rounding off, we get:
\[ \text{Volume} \approx 88 \text{ cm}^3 \]

### Final Answer:
The volume of the solution containing 18.0 g of NaCl is approximately **88 cm³**.