A solution contains `75 mg NaCl` per `mL`. To what extent must it be diluted to give a solution of concentration `15 mg NaCl` per `mL` of solution.

Options

A. 3 times
B. 4 times
C. 5 times
D. 6 times

Explanation

To solve the problem of diluting a solution containing 75 mg NaCl per mL to achieve a concentration of 15 mg NaCl per mL, we can follow these steps:

### Step 1: Understand the Initial and Final Concentrations
- Initial concentration (C1) = 75 mg/mL
- Final concentration (C2) = 15 mg/mL

### Step 2: Set Up the Equation for Dilution
In dilution, the number of moles of solute before dilution is equal to the number of moles of solute after dilution. This can be expressed as:
\[ n_1 = n_2 \]
Where:
- \( n_1 \) = moles before dilution
- \( n_2 \) = moles after dilution

### Step 3: Relate Moles to Concentration and Volume
Moles can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Thus, we can express the moles before and after dilution as:
\[ n_1 = \frac{m_1}{M} \]
\[ n_2 = \frac{m_2}{M} \]
Where:
- \( m_1 = C1 \times V \) (mass before dilution)
- \( m_2 = C2 \times V' \) (mass after dilution)
- \( M \) = molar mass of NaCl = 58 g/mol (or 58000 mg/mol)

### Step 4: Calculate Mass Before and After Dilution
- Mass before dilution:
\[ m_1 = C1 \times V = 75 \, \text{mg/mL} \times V \]
- Mass after dilution:
\[ m_2 = C2 \times V' = 15 \, \text{mg/mL} \times V' \]

### Step 5: Set Up the Equation
Since \( n_1 = n_2 \):
\[ \frac{75 \times V}{58} = \frac{15 \times V'}{58} \]
We can cancel out the molar mass (58) from both sides:
\[ 75 \times V = 15 \times V' \]

### Step 6: Solve for \( V' \)
Rearranging the equation gives:
\[ V' = \frac{75}{15} \times V \]
\[ V' = 5 \times V \]

### Conclusion
To achieve a concentration of 15 mg NaCl per mL from an initial concentration of 75 mg NaCl per mL, the solution must be diluted to **5 times the initial volume**.