Vapor pressure of the solution = Partial pressure of ethanol + partial pressure of water.
Partial pressure of ethanol can be calculated from the molefraction of ethanol and vapor pressure of pure ethanol at [tex] 20^{0}C [/tex]
Density of ethanol = 0.800 g/mL
Finding out mass of ethanol = [tex] 44.0 mL * \frac{0.800 g}{1 mL} = 35.2 g [/tex]
Moles of ethanol = [tex] 35.2 g * \frac{1 mol}{46.07 g} =0.764 mol [/tex]
Density of water = 1.00 g/mL
Mass of water = [tex] 56.0 mL * \frac{1.00 g}{mL} = 56.0 g [/tex]
Moles of water = [tex] 56.0 g * \frac{1 mol}{18.02 g} = 3.11 mol [/tex]
Total number of moles of the solution = 0.764 mol + 3.11 mol = 3.874 mol
Molefraction of ethanol = [tex] \frac{0.764 mol}{3.874 mol} = 0.197 [/tex]
Molefraction of water = [tex] \frac{3.11 mol}{3.874 mol} = 0.803 [/tex]
Vapor pressure of pure ethanol at [tex] 20^{0}C = 59.3 mm Hg [/tex]
Vapor pressure of pure water at [tex] 20^{0}C = 17.5 mmHg [/tex]
Using Raoult's law to find out partial pressures:
[tex] P_{ethanol} = (Molefraction_{ethanol}) P^{0}_{ethanol} [/tex]
[tex] P_{ethanol}= 0.197 (59.3 mmHg) [/tex] = 11.68 mmHg
[tex] P_{water} = (Molefraction)_{water}(P^{0}_{water}) [/tex]
[tex] P_{water} = [/tex][tex] 0.803 (17.5 mmHg) [/tex] = 14.05 mmHg
Total vapor pressure of the solution = 11.68 mmHg + 14.05 mmHg
= 25.73 mmHg
