The pH of an aqueous solution containing 1M benzoic acid ($$pK_a = 4.20$$) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is _________ mL.

Henderson-Hasselbalch equation: $$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$$

$$4.5 = 4.2 + \log\frac{[\text{benzoate}]}{[\text{benzoic acid}]}$$

$$\log\frac{[\text{benzoate}]}{[\text{benzoic acid}]} = 0.3 \Rightarrow \frac{[\text{benzoate}]}{[\text{benzoic acid}]} = 2$$

In 300 mL buffer: let volume of benzoic acid = $$V$$ mL (1M), then volume of sodium benzoate = $$(300 - V)$$ mL (1M).

$$\frac{300 - V}{V} = 2 \Rightarrow 300 - V = 2V \Rightarrow 300 = 3V \Rightarrow V = 100$$ mL.

Therefore, the answer is $$\boxed{100}$$.