$n_{Na}=\frac{2,3}{23}=0,1(mol)$
$m_{NaOH}=\frac{100.10}{100}=10(g)$
$PTHH: 2Na+2H_{2}O→2NaOH+H_{2}↑$
$mol:$ 0,1 → 0,1 → 0,05
$m_{ddSPU}=m_{NaOH}+m_{Na}-m_{H_{2}}=100+2,3-0,05.2=102,2(g)$
$n_{NaOH}=0,1+\frac{10}{40}=0,35(mol)$
C%$_{NaOH}=\frac{0,35.40}{102,2}.100$%$=13,7$%
$V_{dd}=\frac{102,2}{1,05}=97(ml)=0,097(l)$
$C_{M_{NaOH}}=\frac{0,35}{0,097}=3,61M$
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