One mole of `(C_(6)H_(5))_(3)C.OH` dissolved in `1000g` of `100%` sulphuric acid lowers the freezing point of sulphuric acid twice as one mole of `CH_(3)OH` shows in `1000g` of `100%` sulphuric acid. Comment on it associated in sulphuric acid.

Explanation

To solve the problem, we need to analyze the freezing point depression caused by the solutes when dissolved in sulfuric acid. The key points to consider are the number of particles (ions) produced by the solutes when they dissolve and how this affects the freezing point of the solvent.

### Step-by-Step Solution:

1. **Understanding Freezing Point Depression**:
The freezing point depression (\( \Delta T_f \)) is given by the formula:
\[
\Delta T_f = i \cdot K_f \cdot m
...