$BaCl_2 + H_2SO_4 => BaSO_4 + 2HCl$$n_{BaCl_2} = \frac{100.5,2}{100.208} = 0,025$ mol
$m_{H_2SO_4} = \frac{100.1,14.20}{100} = 22,8$ gam
$m_{H_2SO_4 pư} = 0,025.98 = 2,45$ gam
=> $m_{H_2SO_4 dư} = 22,8-2,45 = 20,35$ gam
$n_{BaSO_4} = n_{BaCl_2} = 0,025$ mol
$m_{dd sau} = 100.1,14 + 100 - 0,025.233 = 208,175$ gam
=> $C$% = $9,775$ %