The vapour pressure of fluorobenzene at `t^(@)C` is given by the equation
log `p(mm Hg)=7.0-(1250)/(t+220)`
Calculate the boiling point of the liquid in `.^(@)C` if the external (applied) pressure is `5.26%` more than required for normal boiling point. (log `2=0.3`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the external pressure
The normal boiling point of a liquid is defined at a pressure of 760 mm Hg. The problem states that the external pressure is 5.26% more than this value.

\[ P_{\text{external}} = 760 \, \text{mm Hg} + 0.0526 \times 760 \, \text{mm Hg} \]

Calculating this:

\[ P_{\text{external}} = 760 \, \text{mm Hg} \times (1 + 0.0526) = 760 \, \text{mm Hg} \times 1.0526 \approx 799.986 \, \text{mm Hg} \]

Rounding this value gives:

\[ P_{\text{external}} \approx 800 \, \text{mm Hg} \]

### Step 2: Use the vapor pressure equation
We have the vapor pressure equation given as:

\[ \log P = 7.0 - \frac{1250}{T + 220} \]

Now we substitute \( P_{\text{external}} = 800 \, \text{mm Hg} \) into the equation:

\[ \log 800 = 7.0 - \frac{1250}{T + 220} \]

### Step 3: Calculate \(\log 800\)
Using the logarithmic properties, we can find \(\log 800\):

\[ \log 800 = \log(8 \times 100) = \log 8 + \log 100 = \log 8 + 2 \]

Since \( \log 2 = 0.3 \), we can find \(\log 8\):

\[ \log 8 = \log(2^3) = 3 \cdot \log 2 = 3 \cdot 0.3 = 0.9 \]

Thus,

\[ \log 800 = 0.9 + 2 = 2.9 \]

### Step 4: Substitute \(\log 800\) back into the equation
Now we can substitute \(\log 800\) into the equation:

\[ 2.9 = 7.0 - \frac{1250}{T + 220} \]

### Step 5: Rearranging the equation
Rearranging the equation to isolate \(T\):

\[ \frac{1250}{T + 220} = 7.0 - 2.9 = 4.1 \]

### Step 6: Solve for \(T\)
Now, we can solve for \(T\):

\[ 1250 = 4.1(T + 220) \]

Expanding this gives:

\[ 1250 = 4.1T + 902 \]

Subtracting 902 from both sides:

\[ 1250 - 902 = 4.1T \]

\[ 348 = 4.1T \]

Now, divide both sides by 4.1:

\[ T = \frac{348}{4.1} \approx 84.78 \, ^\circ C \]

### Final Answer
The boiling point of the liquid is approximately \( 84.78 \, ^\circ C \).

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