Question

A mixture of ethanol and 1−propanol behaves ideally at 36°C and is in equilibrium with its vapor. If the mole fraction of ethanol in the solution is 0.65, calculate its mole fraction in the vapor phase at this temperature. (The vapor pressures of pure ethanol and 1−propanol at 36°C are 108 and 40.0 mmHg, respectively.)

Answer 1

Answer:

The mixture behaves ideally ∴ Using Raoults law

P = x × p°

Vapor pressure of a fraction = mole fraction × vapor pressure of pure substance

$P_{ethanol}$ = 0.65 × 108 mmHg

             =  70.2 mmHg

$P_{propanol}$  =  0.35× 40.0 mmHg

                = 4.9 mmHg

then $P_{T}$ = 70.2 + 4.9 = 75.1 mmHg

y = $\frac{P}{P_{T} }$

∴ mole fraction of ethanol = 70.2/ 75.1 = 0.935

mole fraction of propanol = 4.9 / 75.1 = 0.0652

Explanation:

The above reaction behave Ideally, which makes the reaction to obey Raoults law.