To `10 mL` of `1 M BaCl_(2)` solution `5mL` of `0.5 M K_(2)SO_(4)` is added. `BaSO_(4)` is precipitated out. What will happen?

Options

A. Freezing point will increase.
B. Boiling point will increase.
C. Freezing point will lower down.
D. Boiling point will lower down.

Explanation

To solve the problem, we will analyze the reaction that occurs when `BaCl₂` is mixed with `K₂SO₄`, and the implications of the formation of `BaSO₄` precipitate.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- We have `10 mL` of `1 M BaCl₂` solution.
- We also have `5 mL` of `0.5 M K₂SO₄` solution.

2. **Calculate the Moles of Each Reactant:**
- For `BaCl₂`:
\[ \text{Moles of BaCl₂} = \text{Concentration} \times \text{Volume} = 1 \, \text{mol/L} \times 0.010 \, \text{L} = 0.01 \, \text{mol} \]
- For `K₂SO₄`:
\[ \text{Moles of K₂SO₄} = 0.5 \, \text{mol/L} \times 0.005 \, \text{L} = 0.0025 \, \text{mol} \]

3. **Write the Balanced Chemical Equation:**
- The reaction between `BaCl₂` and `K₂SO₄` can be represented as:
\[ BaCl₂ + K₂SO₄ \rightarrow BaSO₄ \downarrow + 2 KCl \]
- Here, `BaSO₄` is the precipitate formed.

4. **Determine the Limiting Reactant:**
- From the balanced equation, 1 mole of `BaCl₂` reacts with 1 mole of `K₂SO₄` to produce 1 mole of `BaSO₄`.
- We have `0.01 mol` of `BaCl₂` and `0.0025 mol` of `K₂SO₄`.
- Since `K₂SO₄` is present in a smaller amount, it is the limiting reactant.

5. **Calculate the Amount of Precipitate Formed:**
- Since `0.0025 mol` of `K₂SO₄` will react completely with `0.0025 mol` of `BaCl₂` to form `0.0025 mol` of `BaSO₄`.
- The amount of `BaSO₄` precipitate formed is `0.0025 mol`.

6. **Effect on Colligative Properties:**
- The formation of a precipitate (`BaSO₄`) is a non-volatile solute.
- According to colligative properties:
- The boiling point of the solution will increase.
- The freezing point of the solution will decrease.

### Conclusion:
- The addition of `K₂SO₄` to `BaCl₂` results in the formation of `BaSO₄` precipitate.
- The boiling point of the solution will increase, and the freezing point will decrease.