If the entropy of vaporisation of a liquid is 110 `JK^(-1)mol^(-1)` and its enthalpy of vaporisation is 50 kJ `mol^(-1)`. The boiling point of the liquid is

Options

A. 354.5 K
B. 454.5 K
C. 554.5 K
D. 445.5 K

Answer

To find the boiling point of the liquid, we can use the relationship between the entropy of vaporization (ΔS) and the enthalpy of vaporization (ΔH) given by the equation:

\[ \Delta S = \frac{\Delta H}{T_b} \]

Where:
- ΔS is the entropy of vaporization,
- ΔH is the enthalpy of vaporization,
- \(T_b\) is the boiling point in Kelvin.

### Step 1: Convert ΔH from kJ/mol to J/mol
Given:
- ΔH = 50 kJ/mol

To convert kJ to J, we multiply by 1000:

\[ \Delta H = 50 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 50000 \, \text{J/mol} \]

### Step 2: Substitute the values into the equation
Now we can substitute the values of ΔS and ΔH into the equation:

\[ \Delta S = 110 \, \text{J/K·mol} \]
\[ \Delta H = 50000 \, \text{J/mol} \]

Substituting these values into the equation:

\[ 110 = \frac{50000}{T_b} \]

### Step 3: Rearranging the equation to solve for \(T_b\)
Rearranging the equation to isolate \(T_b\):

\[ T_b = \frac{50000}{110} \]

### Step 4: Calculate \(T_b\)
Now, we perform the calculation:

\[ T_b = \frac{50000}{110} \approx 454.545 \, \text{K} \]

### Step 5: Rounding off the answer
Rounding off to three significant figures, we get:

\[ T_b \approx 454.5 \, \text{K} \]

Thus, the boiling point of the liquid is approximately **454.5 K**.

### Final Answer:
The boiling point of the liquid is **454.5 K**.

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