\(m_{HCl\left(dd.1\right)}=\dfrac{m_1.45}{100}=0,45.m_1\left(g\right)\)
\(m_{HCl\left(dd.2\right)}=\dfrac{m_2.15}{100}=0,15.m_2\left(g\right)\)
\(C\%_{dd.sau.khi.trộn}=\dfrac{0,45.m_1+0,15.m_2}{m_1+m_2}.100\%=25\%\)
=> \(0,45.m_1+0,15.m_2=0,25.m_1+0,25.m_2\)
=> \(0,2.m_1=0,1.m_2\)
=> m1 : m2 = 1 : 2
