Đáp án:
+/ Tính C%:
$C\%_{KOH}=5,3\%$
+/ Tính $C_M$
$C_M=1M$
Giải thích các bước giải:
$n_K=\dfrac{3,9}{39}=0,1\ mol$
$2K+2H_2O\to 2KOH+H_2$
Theo PTHH: $n_{KOH}=n_{K}=0,1\ mol$
$n_{H_2}=\dfrac{1}{2}.n_K=0,05\ mol$
$m_{dd\ sau\ pư}=m_K+m_{H_2O}-m{H_2}=3,9+101,8-0,05.2=105,6g$
+/ Tính C%:
$m_{KOH}=m_{chất\ tan}=0,1.(39+16+1)=5,6g\\⇒C\%_{KOH}=\dfrac{5,6}{105,6}.100\%=5,3\%$
+/ Tính $C_M$
$V_{dd}=\dfrac{m_{dd}}{d}=\dfrac{105,6}{1,056}=100ml=0,1\ lít$
$C_M=\dfrac{n}{V}=\dfrac{0,1}{0,1}=1M$
