1) Ta có:
\(m_{ddNa2CO3}=\frac{10,6}{10,6\%}=100\left(g\right)\)
\(\Rightarrow m_{H2O}=100-10,6=89,4\left(g\right)\)
2)
Giả sử có 1 mol Na2CO3
\(V_{dd}=\frac{1}{0,2}=5\left(l\right)\)
\(m_{Na2CO3}=1.\left(23.2+60\right)=106\left(g\right)\)
\(\Rightarrow V_{Na2CO3}=106.0,5=53\left(ml\right)\)
\(\Rightarrow V_{H2O}=5000-53=4947\left(ml\right)\)
Hòa tan 106 gam Na2CO3 vào 4947 ml nước