Answer
5.2 M C2H5OH
Work Step by Step
75.0 mL * $\frac{0.79 g}{mL}$ * $\frac{1mol}{46.07 g}$ = 1.3 mol C2H5OH; molarity= $\frac{1.3 mol}{0.250L}$ = 5.2 M C2H5OH
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