Đáp án:
\(m_{\text{dd}\ Al_2{(SO_4)}_3.18H_2O}=133,2\ \text{gam}\)
Giải thích các bước giải:
Theo đề ra, ta có:
\(\dfrac{m_{Al_2{(SO_4)}_3.18H_2O}}{m_{\text{dd}\ Al_2{(SO_4)}_3.18H_2O}}\cdot 100\%=20\%\Leftrightarrow m_{\text{dd}\ Al_2{(SO_4)}_3.18H_2O}=\dfrac{m_{Al_2{(SO_4)}_3.18H_2O}}{20\%}=\dfrac{26,64}{20\%}=133,2\ \text{gam}\)