$Ca ( OH )_2+2 HCl \rightarrow CaCl _2+2 H _2 O$

Volume of $Ca ( OH )_2=10 ml$

Volume of $HCl =20 ml$

Concentration of $HCl =0.5 M$.

No. of milli moles of $HCl =10$

No. of milli moles of $Ca ( OH )_2=5$.

$\text { i.e. }M _{ Ca ( OH )_2}=\frac{\text { no. of milli moles }}{ V ( ml )}=\frac{5}{10}$

$=0.5\,M.$

Reaction with $H _2 SO _4$

$Ca ( OH )_2+ H _2 SO _4 \rightarrow CaSO _4+2 H _2 O$

No. of milli moles of $Ca ( OH )_2=20 \times 0.5$

$=10$

i.e. no. of milli moles of $H _2 SO _4=10$

$\Rightarrow M _{ H _2 so }=\frac{\text { no. of mil lim oles }}{ V ( ml )}$

$=\frac{10}{10}$

$=1\,M$