\(C\%_{KCl}=\dfrac{20}{600}\cdot100\%=3.33\%\)
\(C\%_{K_2SO_4}=\dfrac{75}{1500}\cdot100\%=5\%\)
\(C\%_{NaCl}=\dfrac{15}{15+45}\cdot100\%=25\%\)
\(n_{HCl}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)
\(m_{dd_{HCl}}=7.3+500=507.3\left(g\right)\)
\(C\%_{HCl}=\dfrac{7.3}{507.3}\cdot100\%=1.44\%\)