The boiling point (in `""^@`C) of 0.1 molal aqueous solution of `CuSO_4.5H_2 O `at 1 bar is closest to: [Given: Ebullioscopic (molal boiling point elevation) constant of water, `K_b = 0.512 K Kg mol^(–1)`] :-

Options

A. `100.36`
B. `99.64`
C. `100.10`
D. `99.90`

Explanation

To find the boiling point of a 0.1 molal aqueous solution of CuSO₄·5H₂O, we will use the formula for boiling point elevation. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Molality (m) = 0.1 mol/kg
- Ebullioscopic constant (K_b) of water = 0.512 °C kg/mol
- The dissociation of CuSO₄·5H₂O in water produces 3 ions: Cu²⁺, SO₄²⁻, and 5H₂O (the water of crystallization does not affect the dissociation).

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) is the number of particles the solute dissociates into:
- CuSO₄ dissociates into 1 Cu²⁺ ion and 1 SO₄²⁻ ion.
- Thus, i = 2.

### Step 3: Calculate the boiling point elevation (ΔT_b)
Using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Substituting the values:
\[ \Delta T_b = 2 \cdot 0.512 \, \text{°C kg/mol} \cdot 0.1 \, \text{mol/kg} \]
Calculating:
\[ \Delta T_b = 2 \cdot 0.512 \cdot 0.1 = 0.1024 \, \text{°C} \]

### Step 4: Calculate the new boiling point (T_b)
The boiling point of pure water is 100 °C. Therefore, the new boiling point is:
\[ T_b = 100 \, \text{°C} + \Delta T_b \]
Substituting the value of ΔT_b:
\[ T_b = 100 \, \text{°C} + 0.1024 \, \text{°C} = 100.1024 \, \text{°C} \]

### Step 5: Round to the nearest hundredth
The closest boiling point in the options provided is:
\[ T_b \approx 100.10 \, \text{°C} \]

### Final Answer
The boiling point of the 0.1 molal aqueous solution of CuSO₄·5H₂O at 1 bar is closest to **100.10 °C**.

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