Nồng độ \(CaCl_2\) trong \(CaCl_2.6H_2O\) là:
\(C\%=\dfrac{M_{CaCl_2}}{M_{CaCl_2.6H_2O}}\cdot100\%=\dfrac{111}{219}\cdot100\%=50,68\%\)
Sơ đồ chéo:
\(CaCl_2.6H_2O\) \(m_1\) 50,68 40
40
\(H_2O\) \(m_2\) 0 10,68
\(\Rightarrow\dfrac{m_1}{m_2}=\dfrac{40}{10,68}\) (*)
\(m_{dd}=V\cdot D=10\cdot1,395=13,95g=m_1+m_2\)
\(\Rightarrow m_2=13,95-m_1\) Thay vào (*) ta được:
\(\Rightarrow\dfrac{m_1}{13,95-m_1}=\dfrac{40}{10,68}\Rightarrow m_1=11,01g\)
\(\Rightarrow m_2=13,95-11,01=2,94g\Rightarrow n_{H_2O}=0,163mol\)
\(V_{H_2O}=0,163\cdot22,4=3,65l\)
