Gọi \(m_{ddNaCl.8\%}=x\left(g\right)\)
\(\Rightarrow m_{NaCl.8\%}=8\%x=0,08x\left(g\right)\)
\(m_{NaCl.20\%}=400\times20\%=80\left(g\right)\)
Ta có: \(m_{NaCl.16\%}=0,08x+80\left(g\right)\)
Ta có: \(m_{ddNaCl.16\%}=x+400\left(g\right)\)
\(C\%_{NaCl.16\%}=\frac{0,08x+80}{x+400}\times100\%=16\%\)
\(\Leftrightarrow\frac{0,08x+80}{x+400}=0,16\)
\(\Leftrightarrow0,08x+80=0,16x+64\)
\(\Leftrightarrow16=0,08x\)
\(\Leftrightarrow x=200\)
Vậy \(m_{ddNaCl.8\%}=200\left(g\right)\)
Đúng 0
Bình luận (0)
