Answer:
Since water and ethanol have the same mass, there is an excess of positive volume at a temperature of 20°C
Explanation:
The component 1 refers to water and the component 2 refers to ethanol. The molar volume is
[tex]V_{ethanol} =\frac{m_{ethanol} }{p_{ethanol} } =\frac{46.06}{0.789} =58.37cm^{3} /mol\\V_{water} =\frac{m_{water} }{p_{water} } =\frac{18}{0.998} =18.03cm^{3} /mol[/tex]
The excess volume is
[tex]V_{E} =\frac{x_{1}m_{1}+x_{2}m_{2} }{p_{m} } -x_{1}V_{1} -x_{2} V_{2}[/tex]
Where xn are the mole fraction of water and ethanol, pm is the density of the mixture.
[tex]V_{E} =\frac{(0.5*18)+(0.5*46.06)}{0.913} -(0.5*18.03)-(0.5*58.37)=55.3cm^{3}[/tex]
of the results, since water and ethanol have the same mass, there is an excess of positive volume at a temperature of 20°C
