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 `2.26%` more than required for normal boiling point. (log `2=0.3`)

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Determine the external pressure
The normal boiling point of a liquid occurs at a pressure of 760 mm Hg. Given that the external pressure is 2.26% more than the normal boiling point, we can calculate the external pressure as follows:

\[ \text{External Pressure} = 760 \, \text{mm Hg} + 0.0226 \times 760 \, \text{mm Hg} \]

Calculating this gives:

\[ \text{External Pressure} = 760 + 17.176 = 777.176 \, \text{mm Hg} \]

### Step 2: Use the vapor pressure equation
The vapor pressure of fluorobenzene at temperature \( t \) is given by the equation:

\[ \log p \, (\text{mm Hg}) = 7.0 - \frac{1250}{t + 220} \]

We can substitute the external pressure (777.176 mm Hg) into this equation:

\[ \log 777.176 = 7.0 - \frac{1250}{t + 220} \]

### Step 3: Calculate \( \log 777.176 \)
Using the properties of logarithms, we can approximate:

\[ \log 777.176 \approx 2.890 \]

### Step 4: Substitute and rearrange the equation
Now we substitute this value into the equation:

\[ 2.890 = 7.0 - \frac{1250}{t + 220} \]

Rearranging gives:

\[ \frac{1250}{t + 220} = 7.0 - 2.890 \]

Calculating the right-hand side:

\[ 7.0 - 2.890 = 4.110 \]

So we have:

\[ \frac{1250}{t + 220} = 4.110 \]

### Step 5: Solve for \( t \)
Cross-multiplying gives:

\[ 1250 = 4.110(t + 220) \]

Expanding this:

\[ 1250 = 4.110t + 902.2 \]

Now, isolating \( t \):

\[ 1250 - 902.2 = 4.110t \]

Calculating the left-hand side:

\[ 347.8 = 4.110t \]

Dividing both sides by 4.110:

\[ t = \frac{347.8}{4.110} \approx 84.7 \, ^\circ C \]

### Final Answer
Thus, the boiling point of the liquid at the given external pressure is approximately:

\[ t \approx 84.7 \, ^\circ C \]