In an acid$-$base titration experiment, $0\mathrm{.}1$MNaOH solution was standardized by using $0\mathrm{.}1140g{H}_2{C}_2{O}_4*2H_{2}O$ dissolved in $100ml$ water. This solution was titrated with NaOH solution until its $2^{nd}-$end$-$point $($with $1ml$ phenolphthalein$)$ and the amount of base used in this titration is $19\mathrm{.}2ml.$

$25ml$ of unknown $HCl$ solution was titrated until its end$-$point and $8\mathrm{.}8ml$ standardized $($above$)$ NaOH was used. Calculate $mg$ of $HCl$ in the unknown. $($:$36\mathrm{.}5\frac{g}{m}ol{H}_2{C}_2{O}_4*2H_{2}O=126\mathrm{.}07\frac{g}{m}ol\}$