Ban đầu:
$m_{H_2SO_4}=278.10\%=27,8g$
Gọi x là mol $SO_3$ thêm.
$SO_3+H_2O\to H_2SO_4$
$\Rightarrow n_{H_2SO_4}=x$
Sau phản ứng:
$m_{H_2SO_4}=27,8+98x$
$m_{dd}=278+80x$
$C\%=19\%$
$\Rightarrow \dfrac{27,8+98x}{278+80x}=0,19$
$\Leftrightarrow 27,8+98x=0,19(278+80x)$
$\Leftrightarrow x=0,3$
$V=22,4x=6,72l$