1. Write the reaction

$\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}$

Sulfuric acid is dibasic, so its normality is:

$N = M \times 2$

for acid-base reaction.

2. Calculate initial equivalents of $\mathrm{H_2SO_4}$

Given:

• Volume $= 100\,\text{mL} = 0.1\,\text{L}$
• Molarity $= 0.1\,\text{M}$

Moles of $\mathrm{H_2SO_4}$:

$0.1 \times 0.1 = 0.01\,\text{mol}$

Since 1 mole of $\mathrm{H_2SO_4}$ gives 2 equivalents:

$\text{equivalents of } \mathrm{H_2SO_4} = 0.01 \times 2 = 0.02$

3. Calculate initial equivalents of $\mathrm{NaOH}$

Given:

• Volume $= 50\,\text{mL} = 0.05\,\text{L}$
• Molarity $= 0.1\,\text{M}$

Moles of $\mathrm{NaOH}$:

$0.1 \times 0.05 = 0.005\,\text{mol}$

For $\mathrm{NaOH}$, 1 mole = 1 equivalent, so:

$\text{equivalents of } \mathrm{NaOH} = 0.005$

4. Find excess acid equivalents after neutralization

$\text{Excess acid equivalents} = 0.02 - 0.005 = 0.015$

5. Total volume after mixing

$100\,\text{mL} + 50\,\text{mL} = 150\,\text{mL} = 0.15\,\text{L}$

6. Normality of remaining $\mathrm{H_2SO_4}$

$N = \frac{\text{equivalents}}{\text{volume in L}} = \frac{0.015}{0.15} = 0.1\,\text{N}$

Now,

$0.1\,\text{N} = 1 \times 10^{-1}\,\text{N}$

So the required integer is:

$\boxed{1}$