a) Ta có:
\({V_{{C_2}{H_5}OH}} = 1200.35\% = 420{\text{ ml}}\)
b) Ta có:
\({V_{{C_2}{H_5}OH}} = 1500.42\% = 630{\text{ ml}} \to {{\text{V}}_{dd{\text{ }}{{\text{C}}_2}{H_5}OH{\text{ 3}}{{\text{0}}^o}}} = \frac{{630}}{{30\% }} = 2100{\text{ ml = 2}}{\text{,1 lít}}\)