Xét $100g$ dung dịch $NaOH$
$n_{NaOH}=\dfrac{100.40\%}{40}=1(mol)$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\to n_{H_2SO_4}=n_{Na_2SO_4}=0,5(mol)$
$\to m_{dd\text{spu}}=0,5.142:35,5\%=200g$
$\to m_{dd H_2SO_4}=200-100=100g$
Vậy $C\%_{H_2SO_4}=\dfrac{0,5.98.100}{100}=49\%$