\(n_{H_2SO_4}=40\%.50=20\left(g\right)\\ Gọi:a=m_{H_2O\left(thêm\right)}\\ \Rightarrow\dfrac{20}{50+a}.100\%=10\%\\ \Leftrightarrow a=150\left(g\right)\\ \Rightarrow V_{H_2O\left(thêm\right)}=\dfrac{150}{1}=150\left(ml\right)\)
$m_{CT_{H_2SO_4}}=50.40\%=20(g)$
$m_{dd_{H_2SO_4(10\%)}}=\dfrac{20}{10\%}=200(g)$
$\to m_{H_2O\text{ thêm}}=200-50=150(g)$
$\to V_{H_2O\text{ thêm}}=\dfrac{150}{1}=150(ml)$