a)
Ta có:
\({V_{{C_2}{H_5}OH}} = 80.90\% = 72{\text{ lít}}\)
\( \to {V_{rượu{\text{ 4}}{{\text{0}}^o}}} = \frac{{72}}{{40\% }} = 180{\text{ lít}}\)
\( \to {V_{{H_2}O}} = 180 - 80 = 100{\text{ lít}}\)
Cần pha thêm 100 lít \(H_2O\) vào dung dịch.
b)
Ta có:
\({V_{{C_2}{H_5}OH}} = 200.60\% = 120{\text{ ml}}\)
\( \to {V_{rượu{\text{ 2}}{{\text{0}}^o}}} = \frac{{120}}{{20\% }} = 600{\text{ ml}}\)
\( \to {V_{{H_2}O}} = 600 - 120 = 480{\text{ ml}}\)
Cần pha thêm 480 ml \(H_2O\).
c)
Ta có:
\({V_{{C_2}{H_5}OH}} = 250.45\% = 112,5{\text{ ml}}\)
\( \to {V_{rượu{\text{ 3}}{{\text{0}}^o}}} = \frac{{112,5}}{{30\% }} = 375{\text{ ml}}\)
\( \to {V_{{H_2}O}} = 375 - 250 = 125{\text{ ml}}\)
Cần pha thêm 125 ml \(H_2O\).