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 34.0 g of the salt will be

Options

A. 120`cm^3`
B. 167`cm^3`
C. 140`cm^3`
D. None

Explanation

To solve the problem, we need to find the volume of the salt solution that contains 34.0 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. **Identify the Given Information:**
- Density of the solution (d) = 1.13 g/cm³
- Percentage of NaCl by weight = 18%
- Mass of NaCl (m_NaCl) = 34.0 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 formula:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.13 \times V \text{ g} \]

3. **Relate the Mass of NaCl to the Mass of the Solution:**
- Since the solution contains 18% NaCl by weight, we can express this as:
\[ \text{Mass of NaCl} = \frac{18}{100} \times \text{Mass of solution} \]
- Substituting the mass of the solution:
\[ 34.0 = \frac{18}{100} \times (1.13 \times V) \]

4. **Rearranging the Equation:**
- Rearranging the equation to solve for \( V \):
\[ 34.0 = 0.18 \times 1.13 \times V \]
\[ V = \frac{34.0}{0.18 \times 1.13} \]

5. **Calculating the Volume:**
- Calculate the denominator:
\[ 0.18 \times 1.13 = 0.2034 \]
- Now, calculate \( V \):
\[ V = \frac{34.0}{0.2034} \approx 167.0 \text{ cm}^3 \]

6. **Final Answer:**
- The volume of the solution containing 34.0 g of NaCl is approximately **167 cm³**.