Calculate the b.pt. of 1 molar aqueous solution of KBr. Given that the density of solution is `1.06 gmL^(-),K_(b)`, for water is 0.52 K kg `"mol"^(-)`[Atomic mass of K = 39. Br=80]

Explanation

To calculate the boiling point of a 1 molar aqueous solution of KBr, we will follow these steps:

### Step 1: Calculate the molar mass of KBr
- The molar mass of K (Potassium) = 39 g/mol
- The molar mass of Br (Bromine) = 80 g/mol
- Therefore, the molar mass of KBr = 39 + 80 = 119 g/mol

### Step 2: Determine the mass of KBr in the solution
- Since the solution is 1 molar, we have 1 mole of KBr.
- Mass of KBr = 1 mole × 119 g/mol = 119 g

### Step 3: Calculate the mass of the solution
- Volume of the solution = 1 L = 1000 mL
- Density of the solution = 1.06 g/mL
- Mass of the solution = Volume × Density = 1000 mL × 1.06 g/mL = 1060 g

### Step 4: Calculate the mass of the solvent (water)
- Mass of solvent = Mass of solution - Mass of solute (KBr)
- Mass of solvent = 1060 g - 119 g = 941 g

### Step 5: Convert the mass of the solvent to kg
- Mass of solvent in kg = 941 g / 1000 = 0.941 kg

### Step 6: Calculate the molality of the solution
- Molality (m) = moles of solute / mass of solvent (kg)
- Molality = 1 mol / 0.941 kg = 1.0626 m

### Step 7: Determine the van 't Hoff factor (i) for KBr
- KBr dissociates into K⁺ and Br⁻ ions.
- Therefore, the number of particles after dissociation (i) = 2.

### Step 8: Calculate the boiling point elevation (ΔTb)
- The formula for boiling point elevation is ΔTb = i × Kb × m
- Where Kb for water = 0.52 K kg/mol
- ΔTb = 2 × 0.52 K kg/mol × 1.0626 m = 1.1051 °C

### Step 9: Calculate the boiling point of the solution
- The boiling point of pure water = 100 °C
- Boiling point of the solution = 100 °C + ΔTb
- Boiling point of the solution = 100 °C + 1.1051 °C = 101.1051 °C

### Final Answer
The boiling point of a 1 molar aqueous solution of KBr is **101.1051 °C**.

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