Heat of solution of `BaCl_(2).2H_(2)O=200 kJ mol^(-1)`
Heat of hydration of `BaCl_(2)=-150 kJ mol^(-1)`
Hence heat of solution of `BaCl_(2)` is

Options

A. 350 kJ
B. 50 kJ
C. `-350 kJ`
D. None of these

Explanation

To find the heat of solution of barium chloride (BaCl₂), we can use the given data about the heat of solution and the heat of hydration. Here’s a step-by-step solution:

### Step 1: Understand the Definitions
- **Heat of Solution (ΔH_solution)**: This is the amount of heat absorbed or released when one mole of a solute dissolves in a solvent.
- **Heat of Hydration (ΔH_hydration)**: This is the amount of heat released when one mole of ions undergoes hydration in water.

### Step 2: Write Down the Given Values
- Heat of solution of BaCl₂·2H₂O = 200 kJ/mol
- Heat of hydration of BaCl₂ = -150 kJ/mol

### Step 3: Set Up the Relationship
The heat of solution can be expressed in terms of the heat of hydration and the heat of formation of the hydrated salt. The relationship can be given as:
\[ \Delta H_{solution} = \Delta H_{hydration} + \Delta H_{formation} \]

### Step 4: Rearranging the Equation
From the above relationship, we can rearrange it to find the heat of solution:
\[ \Delta H_{solution} = \Delta H_{hydration} + \Delta H_{formation} \]
We need to find the heat of solution of BaCl₂, which can be calculated as:
\[ \Delta H_{solution} = \Delta H_{hydration} + \Delta H_{formation} \]

### Step 5: Substitute the Values
Now we can substitute the known values into the equation:
- We know that the heat of hydration is -150 kJ/mol.
- The heat of solution of BaCl₂·2H₂O is 200 kJ/mol.

Thus, we can write:
\[ \Delta H_{solution} = 200 \text{ kJ/mol} - (-150 \text{ kJ/mol}) \]

### Step 6: Calculate the Heat of Solution
Now, perform the calculation:
\[ \Delta H_{solution} = 200 \text{ kJ/mol} + 150 \text{ kJ/mol} \]
\[ \Delta H_{solution} = 350 \text{ kJ/mol} \]

### Step 7: Conclusion
The heat of solution of barium chloride (BaCl₂) is **350 kJ/mol**.