$20 \mathrm{~mL}$ of calcium hydroxide was consumed when it was reacted with $10 \mathrm{~mL}$ of unknown solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$. Also $20 \mathrm{~mL}$ standard solution of $0.5 ~\mathrm{M} ~\mathrm{HCl}$ containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of $35.5 \mathrm{~mL}$ whereas the burette showed $25.5 \mathrm{~mL}$ initially. The concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is _____________ M. (Nearest integer)

Reaction with HCl:<br/><br/> $$\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}$$ <br/><br/> Volume of $\mathrm{Ca}(\mathrm{OH})_2=10 \mathrm{ml}$<br/><br/> Volume of $\mathrm{HCl}=20 \mathrm{ml}$<br/><br/> Concentration of $\mathrm{HCl}=0.5 \mathrm{M}$. <br/><br/> No. of milli moles of $\mathrm{HCl}=10$<br/><br/> No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=5$. <br/><br/> i.e. $\mathrm{M}_{\mathrm{Ca}(\mathrm{OH})_2}=\frac{\text { no. of milli moles }}{\mathrm{V}(\mathrm{ml})}=\frac{5}{10}$ $=0.5 \mathrm{M}$. <br/><br/> Reaction with $\mathrm{H}_2 \mathrm{SO}_4$:<br/><br/> $$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O} \text {. }$$ <br/><br/> No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=20 \times 0.5$ $=10$<br/><br/> i.e. no. of milli moles of $\mathrm{H}_2 \mathrm{SO}_4=10$ <br/><br/> $$ \begin{aligned} \Rightarrow & \mathrm{M}_{\mathrm{H}_2 \mathrm{SO}_4}=\frac{\text { no, of milli moles }}{\mathrm{V}(\mathrm{ml})} \\\\ & =\frac{10}{10} \\\\ & =1 \mathrm{M} \end{aligned} $$ <br/><br/> So, the concentration of $\mathrm{H}_2 \mathrm{SO}_4$ is 1 M.