At 25°C, the vapor pressure of benzene is 96.0 mm Hg; 30.3 mm Hg for toluene, and 23.76 mm hg).
1. What is:
(i) the vapor pressure of a mixture of 10.0 g of benzene and 20.0 g toluene at 25 oC, and;
(ii) the mole fraction benzene and toluene in the vapor of the solution?
2. What is the vapor pressure of a solution of 15.0 g of NaNO3 in 50.0 g of water at 25°
Please answer all questions
Thank You
As per Raoults law
Pa = Xa x Pao
mixture of 10.0 g of benzene and 20.0 g toluene
10/78g/mol = 0.128 mole
20 /92g/mol = 0.217 mole
So mole fraction of benzene = 0.128/(0.128 + 0.217) = 0.37
mole fraction of toluene = 0.217/(0.128 + 0.217) = 0.628
Total vapour pressure = 96 x 0.37 + 30.3 x 0.628
= 54.55 mm Hg
b) mole fraction of benzene in vapour is 96 x 0.37/54.55 = 0.651
mole fraction of toluene in vapour = 30.3 x 0.628/54.55 = 0.349
c) 15 g of NaNO3 = 15/85g/mol = 0.176 moles
50 g of water is 50/18g/mol = 2.778 moles
mole fraction of water = 2.778/(2.778 + 0.176) = 0.94
Vapour pressure of the solution = 0.94 x 23.76 = 22.34 mm Hg
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