\(C_M\)\(_{HCl}\)\(_{18,25\%}\)=\(\dfrac{10.1,2.18,25}{36,5}\)=6(M) (1)
\(C_M\)\(_{HCl}\)\(_{13\%}\)=\(\dfrac{10.1,123.13}{36,5}\)\(\simeq\)4(M) (2)
Đặt \(V_{HCl\left(1\right)}\)=x(l)
\(V_{HCl\left(2\right)}\)=y(l)
Ta có: \(\dfrac{n_{HCl\left(1\right)}}{V_{HCl\left(1\right)}}\)= 6
\(\dfrac{n_{HCl\left(2\right)}}{V_{HCl\left(2\right)}}\)= 4
Suy ra: \(n_{HCl\left(1\right)}\)=6x(mol)
\(n_{HCl\left(2\right)}\)=4x(mol)
Do đó: \(C_M\)\(_{HCl}\)\(_{4,5M}\)=\(\dfrac{6x+4y}{x+y}\)=4,5
\(\Rightarrow\)\(6x+4y\)=\(4,5x+4,5y\)
\(\Rightarrow\)\(1,5x\)=\(0,5y\)
\(\Rightarrow\dfrac{x}{y}\)=\(\dfrac{1}{3}\)
Vậy tỉ lệ thể tích dung dịch là 1 : 3 để pha thành dung dịch HCl 4,5M
