Hóa học trộn 400ml NaOH 0,5M với 200 ml NaOH 20% (D = 1,25 g/ml) 10/09/2021 By Ruby trộn 400ml NaOH 0,5M với 200 ml NaOH 20% (D = 1,25 g/ml)
– 400ml dd NaOH: $n_{NaOH}=0,4.0,5=0,2(mol)$ – 200ml dd NaOH: $m_{dd}=200.1,25=250g$ $\Rightarrow n_{NaOH}=\dfrac{250.20\%}{40}=1,25(mol)$ – Sau khi trộn: $n_{NaOH}=1,25+0,2=1,45(mol)$ $V_{dd}=0,6l$ $\to C_{M_{NaOH}}=\dfrac{1,45}{0,6}=2,4M$ Trả lời
– 400ml dd NaOH:
$n_{NaOH}=0,4.0,5=0,2(mol)$
– 200ml dd NaOH:
$m_{dd}=200.1,25=250g$
$\Rightarrow n_{NaOH}=\dfrac{250.20\%}{40}=1,25(mol)$
– Sau khi trộn:
$n_{NaOH}=1,25+0,2=1,45(mol)$
$V_{dd}=0,6l$
$\to C_{M_{NaOH}}=\dfrac{1,45}{0,6}=2,4M$