1.
$m_{H_2SO_4}=100.9,8\%=9,8g$
Gọi x là $m_{H_2O}$ thêm.
$m_{dd}=100+x(g)$
$\Rightarrow \dfrac{9,8.100}{100+x}=4,9$
$\Leftrightarrow x=100(g)$
2.
$n_{SO_3}=\dfrac{8}{80}=0,1 mol$
$SO_3+H_2O\to H_2SO_4$
$\Rightarrow n_{H_2SO_4}=0,1 mol$
$m_{H_2SO_4}=0,1.98=9,8g$
$\Rightarrow m_{dd}=9,8:19,6\%=50g$
BTKL: $m_{H_2O}=50-8=42g$
3.
Gọi x là số mol $Na_2O$ thêm.
$Na_2O+H_2O\to 2NaOH$
$\Rightarrow n_{NaOH}=2x$
$m_{dd}=96+62x(g)$
$\Rightarrow \dfrac{40.2x.100}{96+62x}=4$
$\Leftrightarrow x=0,05$
$\Rightarrow m_{Na_2O}=62x=3,1g$
4.
$m_{H_2SO_4}=250.20\%=50g$
$\Rightarrow m_{dd 74\%}=50:74\%=67,57g$
$\Rightarrow V_{dd}=67,57:1,664=40,6ml$
