1. Find the concentration of the calcium hydroxide solution

The burette contains $\mathrm{Ca(OH)_2}$.

In standardization:

• Initial burette reading $= 25.5\,\mathrm{mL}$
• Final burette reading $= 35.5\,\mathrm{mL}$

So, volume of $\mathrm{Ca(OH)_2}$ used is $V_{\mathrm{Ca(OH)_2}} = 35.5-25.5 = 10.0\,\mathrm{mL} = 0.010\,\mathrm{L}$

Given:

• Volume of HCl $= 20\,\mathrm{mL} = 0.020\,\mathrm{L}$
• Molarity of HCl $= 0.5\,\mathrm{M}$

Moles of HCl: $n_{\mathrm{HCl}} = M V = 0.5 \times 0.020 = 0.010\,\mathrm{mol}$

Reaction: $\mathrm{Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O}$

Thus, moles of $\mathrm{Ca(OH)_2}$ required are $n_{\mathrm{Ca(OH)_2}} = \frac{0.010}{2} = 0.005\,\mathrm{mol}$

These $0.005$ mol are present in $0.010\,\mathrm{L}$, so molarity of $\mathrm{Ca(OH)_2}$ is $M_{\mathrm{Ca(OH)_2}} = \frac{0.005}{0.010} = 0.5\,\mathrm{M}$

2. Use this to find concentration of unknown $\mathrm{H_2SO_4}$

Given:

• $20\,\mathrm{mL}$ of $\mathrm{Ca(OH)_2}$ consumed with $10\,\mathrm{mL}$ of unknown $\mathrm{H_2SO_4}$

So, $V_{\mathrm{Ca(OH)_2}} = 20\,\mathrm{mL} = 0.020\,\mathrm{L}$ $M_{\mathrm{Ca(OH)_2}} = 0.5\,\mathrm{M}$

Moles of $\mathrm{Ca(OH)_2}$ used: $n_{\mathrm{Ca(OH)_2}} = 0.5 \times 0.020 = 0.010\,\mathrm{mol}$

Reaction with sulfuric acid: $\mathrm{Ca(OH)_2 + H_2SO_4 \rightarrow CaSO_4 + 2H_2O}$

This is a $1:1$ reaction, so $n_{\mathrm{H_2SO_4}} = 0.010\,\mathrm{mol}$

Volume of $\mathrm{H_2SO_4}$ solution used: $V_{\mathrm{H_2SO_4}} = 10\,\mathrm{mL} = 0.010\,\mathrm{L}$

Therefore, molarity of $\mathrm{H_2SO_4}$ is $M_{\mathrm{H_2SO_4}} = \frac{0.010}{0.010} = 1.0\,\mathrm{M}$

3. Nearest integer

$1.0 \approx 1$

So, the required integer answer is 1.