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Câu 1

$a/ m_{ct}= n.M=0,1.40=4g$

$c/ m_{dd}= \dfrac{m_{ct}}{C\%}= \dfrac{4}{5\%} = 80g$

$b/ m_{nc}= m_{dd}-m_{ct}= 80-4= 76g$

$d/ n_{HCl}= \dfrac{m}{M}= \dfrac{7,3}{36,5}=\dfrac{1}{5}mol$

$e/ m_{dd}= m_{ct}+m_{H_2O}=7,3+192,7=200g$

$g/ C\%= \dfrac{m_{ct}.100}{m_{dd}}=\dfrac{7,3.100}{200}=3,65\%$

$i/ m_{ct}=m_{dd}-m_{nc}= 205,3-200=5,3g$

$h/ n_{Na_2CO_3}=\dfrac{m}{M}=\dfrac{5,3}{106}=0,05mol$

$k/ C\%=\dfrac{m_{ct}.100}{m_dd}=\dfrac{5,3.100}{205,3}=2,58\%$

$n/ m_{dd}=\dfrac{m_{ct}}{C\%}=\dfrac{17,1}{6,84\%}=250g$

$l/ n_{BaO}=\dfrac{m}{M}=\dfrac{17,1}{153}=0,111mol$

$m/ m_{nc}=m_{dd}-m_{ct}=250-17,1=232,9g$