Ta có: $m_{NaOH}$/dd NaOH 8% = 100 × 8% = 8 (g)
Gọi $m_{dd NaOH cần}$ = a (g)
⇒ $m_{NaOH}$/dd NaOH 20% = a × 20% = 0.2a (g)
$C%_{dd NaOH mới}$ = $\frac{8 + 0.2a}{100 + a}$ × 100% = 17.5%
⇒ 8 + 0.2a = 17.5% × ( 100 + a )
⇒ 0.025a = 9.5 ⇒ a = 380 (g)
Vậy $m_{dd NaOH cần}$ = 380 g
~ GỬI BẠN ~