\(n_{HCl}=\dfrac{30.7,3\%}{36,5}=0,06\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ n_{KOH}=n_{HCl}=0,06\left(mol\right)\\ m_{KOH}=0,06.56=3,36\left(g\right)\\ m_{ddKOH}=\dfrac{3,36.100}{5}=67,2\left(g\right)\\ V_{ddKOH}=\dfrac{67,2}{1,082}\approx\text{62,1072089(ml)}\)