What will be the boliling point of bromine when `174.5 mg` of octa-atomic sulphur is added to `78 g` of bromine? `k'_(b)` for `Br_(2)` is `5.2 Kmol^(-1) kg` and `b.pt.` of `Br_(2) " is " 332.15 K`

Options

A. 415.23 K
B. 310 K
C. 220.92 K
D. 332.19 K

Answer

To find the boiling point of bromine when 174.5 mg of octa-atomic sulfur (S8) is added to 78 g of bromine (Br2), we will follow these steps:

### Step 1: Convert the mass of the solute (S8) from milligrams to grams.
Given:
- Mass of S8 = 174.5 mg

To convert milligrams to grams:
\[ \text{Mass of S8 (in grams)} = \frac{174.5 \text{ mg}}{1000} = 0.1745 \text{ g} \]

### Step 2: Calculate the molar mass of octa-atomic sulfur (S8).
The atomic mass of sulfur (S) is approximately 32 g/mol. Therefore, the molar mass of S8 is:
\[ \text{Molar mass of S8} = 8 \times 32 \text{ g/mol} = 256 \text{ g/mol} \]

### Step 3: Calculate the mass of the solvent (bromine) in kilograms.
Given:
- Mass of Br2 = 78 g

To convert grams to kilograms:
\[ \text{Mass of Br2 (in kg)} = \frac{78 \text{ g}}{1000} = 0.078 \text{ kg} \]

### Step 4: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
First, calculate the number of moles of S8:
\[ \text{Moles of S8} = \frac{\text{Mass of S8}}{\text{Molar mass of S8}} = \frac{0.1745 \text{ g}}{256 \text{ g/mol}} \approx 0.000682 \text{ mol} \]

Now, calculate the molality:
\[ \text{Molality (m)} = \frac{\text{Moles of S8}}{\text{Mass of Br2 (in kg)}} = \frac{0.000682 \text{ mol}}{0.078 \text{ kg}} \approx 0.00874 \text{ mol/kg} \]

### Step 5: Calculate the elevation in boiling point (ΔTb).
Using the formula:
\[ \Delta T_b = K_b \times m \]
where \( K_b \) for bromine is given as 5.2 K·kg/mol.

Now, substituting the values:
\[ \Delta T_b = 5.2 \text{ K·kg/mol} \times 0.00874 \text{ mol/kg} \approx 0.0454 \text{ K} \]

### Step 6: Calculate the boiling point of the solution.
The boiling point of pure bromine (Br2) is given as 332.15 K. Therefore, the boiling point of the solution is:
\[ \text{Boiling point of solution} = \Delta T_b + \text{Boiling point of solvent} \]
\[ \text{Boiling point of solution} = 0.0454 \text{ K} + 332.15 \text{ K} \approx 332.1954 \text{ K} \]

### Final Answer:
The boiling point of the solution is approximately **332.195 K**.

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