The normality of $$H_2SO_4$$ in the solution obtained on mixing 100 mL of 0.1 M $$H_2SO_4$$ with 50 mL of 0.1 M NaOH is _____ $$\times 10^{-1}$$ N.

We need to find the normality of $$H_2SO_4$$ in the solution obtained by mixing 100 mL of 0.1 M $$H_2SO_4$$ with 50 mL of 0.1 M NaOH.

Initially, the millimoles of $$H_2SO_4$$ in 100 mL of 0.1 M $$H_2SO_4$$ is calculated as $$100 \times 0.1 = 10$$ mmol, and the millimoles of NaOH in 50 mL of 0.1 M NaOH is $$50 \times 0.1 = 5$$ mmol.

Since $$H_2SO_4$$ is a diprotic acid, it reacts with NaOH according to the equation

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

Thus, 5 mmol of NaOH will react with $$\frac{5}{2} = 2.5$$ mmol of $$H_2SO_4$$.

After the reaction, the remaining $$H_2SO_4$$ is $$10 - 2.5 = 7.5$$ mmol.

The total volume of the mixture becomes $$100 + 50 = 150$$ mL. The remaining $$H_2SO_4$$ still has both acidic protons available, so its milliequivalents are $$7.5 \times 2 = 15$$ meq (since the n-factor of $$H_2SO_4$$ is 2).

$$\text{Normality} = \frac{15}{150} = 0.1 \text{ N} = 1 \times 10^{-1} \text{ N}$$

Hence, the answer is 1.