Trong V ml dung dịch $K_2CO_3$ 10% có D = 1,25(g/ml), ta có :
$m_{dd} = 1,25V(gam)$
$m_{K_2CO_3} = 1,25V.10\% = 0,125V(gam)$
Sau khi trộn :
$m_{dd} = 1,25V + 150(gam)$
$m_{K_2CO_3} = 0,125V + 150.4\% = 0,125V + 6(gam)$
Suy ra :
$\dfrac{0,125V + 6}{1,25V + 150} = \dfrac{6,4}{100}$
$\Rightarrow V = 80(ml)$