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 26.0 g of the salt will be

Options

A. 127.8`cm^3`
B. 120`cm^3`
C. 12.8`cm^3`
D. 50`cm^3`

Explanation

To solve the problem, we need to find the volume of the salt solution that contains 26.0 g of NaCl, given that the density of the solution is 1.13 g/cm³ and it is 18% NaCl by weight.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Density of the solution (d) = 1.13 g/cm³
- Percentage of NaCl by weight = 18%
- Mass of NaCl (salt) = 26.0 g

2. **Set Up the Equation for Mass of the Solution:**
Let the volume of the solution be \( V \) cm³. The mass of the solution can be calculated using the density formula:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.13 \, \text{g/cm}^3 \times V \, \text{cm}^3 = 1.13V \, \text{g} \]

3. **Calculate the Mass of NaCl in Terms of the Mass of the Solution:**
Since the solution contains 18% NaCl by weight, we can express the mass of NaCl as:
\[ \text{Mass of NaCl} = 0.18 \times \text{Mass of solution} \]
Substituting the mass of the solution:
\[ 26.0 \, \text{g} = 0.18 \times (1.13V) \]

4. **Solve for Volume \( V \):**
Rearranging the equation gives:
\[ 26.0 = 0.18 \times 1.13V \]
\[ V = \frac{26.0}{0.18 \times 1.13} \]
Now, calculate the denominator:
\[ 0.18 \times 1.13 = 0.2034 \]
Therefore:
\[ V = \frac{26.0}{0.2034} \approx 127.8 \, \text{cm}^3 \]

5. **Final Result:**
The volume of the solution containing 26.0 g of NaCl is approximately **127.8 cm³**.