When 600 mL of 0.2 M $$HNO_3$$ is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is _____ $$\times 10^{-2}$$ °C (Enthalpy of neutralisation = 57 kJ mol$$^{-1}$$ and Specific heat of water = 4.2 J K$$^{-1}$$ g$$^{-1}$$) (Neglect heat capacity of flask)

We need to find the rise in temperature when 600 mL of 0.2 M $$HNO_3$$ is mixed with 400 mL of 0.1 M NaOH.

First, we find the moles of acid and base. Moles of $$HNO_3 = 0.6 \times 0.2 = 0.12$$ mol. Moles of NaOH $$= 0.4 \times 0.1 = 0.04$$ mol.

Since NaOH is the limiting reagent, the number of moles neutralised is 0.04 mol. The heat released during neutralisation is:

$$q = n \times \Delta H_{neutralisation} = 0.04 \times 57 \times 10^3 = 2280 \text{ J}$$

Now, the total volume of the mixed solution is $$600 + 400 = 1000$$ mL. Assuming the density of the solution is approximately 1 g/mL, the total mass is $$m = 1000$$ g.

Using $$q = m \times c \times \Delta T$$, where $$c = 4.2$$ J K$$^{-1}$$ g$$^{-1}$$:

$$\Delta T = \frac{q}{m \times c} = \frac{2280}{1000 \times 4.2} = \frac{2280}{4200} = 0.5429 \text{ °C}$$

Expressing this as $$\times 10^{-2}$$ °C: $$\Delta T = 54.29 \times 10^{-2}$$ °C.

Rounding to the nearest integer, the answer is $$54 \times 10^{-2}$$ °C.

Hence, the correct answer is 54.