Bromine is prepared commercially by the reaction :
`2Br^(-)(aq)+Cl_(2)(aq)rarr 2Cl^(-)(aq)+Br_(2)(aq)`
Suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.050 M solution of `Cl_(2)` is needed to react completely the `Br^(-)`?

Explanation

To solve the problem, we need to determine how much volume of chlorine gas (Cl₂) is required to react completely with the bromide ions (Br⁻) present in the sodium bromide (NaBr) solution.

### Step-by-step Solution:

1. **Calculate the moles of Br⁻ in the NaBr solution:**
- We are given a 50.0 mL solution of 0.060 M NaBr.
- To find the number of moles of Br⁻, we use the formula:
\[ \text{Moles of Br}^- = \text{Molarity} \times \text{Volume (in L)} \]
- Convert 50.0 mL to liters:
\[ 50.0 \, \text{mL} = \frac{50.0}{1000} = 0.0500 \, \text{L} \]
- Now calculate the moles:
\[ \text{Moles of Br}^- = 0.060 \, \text{mol/L} \times 0.0500 \, \text{L} = 0.0030 \, \text{mol} \]

2. **Determine the moles of Cl₂ required:**
- From the balanced chemical equation:
\[ 2 \, \text{Br}^- + \text{Cl}_2 \rightarrow 2 \, \text{Cl}^- + \text{Br}_2 \]
- We see that 2 moles of Br⁻ react with 1 mole of Cl₂. Therefore, the moles of Cl₂ required can be calculated as follows:
\[ \text{Moles of Cl}_2 = \frac{1}{2} \times \text{Moles of Br}^- = \frac{1}{2} \times 0.0030 \, \text{mol} = 0.0015 \, \text{mol} \]

3. **Calculate the volume of Cl₂ solution needed:**
- We are given a 0.050 M solution of Cl₂. To find the volume needed to obtain 0.0015 moles of Cl₂, we use the formula:
\[ \text{Volume (in L)} = \frac{\text{Moles}}{\text{Molarity}} \]
- Substitute the values:
\[ \text{Volume of Cl}_2 = \frac{0.0015 \, \text{mol}}{0.050 \, \text{mol/L}} = 0.030 \, \text{L} \]
- Convert the volume from liters to milliliters:
\[ 0.030 \, \text{L} = 0.030 \times 1000 = 30.0 \, \text{mL} \]

### Final Answer:
The volume of 0.050 M Cl₂ solution needed to react completely with the Br⁻ is **30.0 mL**.