The vapor pressures of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and 1 -propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) at \(35^{\circ} \mathrm{C}\) are \(100 \mathrm{mmHg}\) and \(37.6 \mathrm{mmHg},\) respectively. Assume ideal behavior and calculate the partial pressures of ethanol and 1 -propanol at \(35^{\circ} \mathrm{C}\) over a solution of ethanol in I-propanol, in which the mole fraction of ethanol is 0.300

Short Answer

Expert verified
The partial pressures of ethanol and 1-propanol at \(35^{\circ} C\) over a solution of ethanol in 1-propanol, in which the mole fraction of ethanol is 0.300, are \(30.0 mmHg \) and \(26.32 mmHg\), respectively.

Step by step solution

01

Calculate mole fraction of 1-propanol

We begin by calculating the mole fraction of 1-propanol in the solution. Since the sum of the mole fractions of all the components in a solution equals to 1, the mole fraction of 1-propanol \(X_{C_{3}H_{7}OH}\) can be calculated as \(1 - 0.300 = 0.700\)
02

Calculate partial pressure of ethanol

Next, we calculate the partial pressure of ethanol using Raoult's Law. According to Raoult's Law, the partial pressure of a component in an ideal mixture is the product of the mole fraction of the component and the vapour pressure of the pure component. Here, \(P_{C_{2}H_{5} OH} = X_{C_{2} H_{5} OH} . P^0_{C_{2}H_{5} OH} = 0.300 . 100 mmHg = 30.0 mmHg\)
03

Calculate partial pressure of 1-propanol

Finally, we calculate the partial pressure of 1-propanol using Raoult's Law. For 1-propanol, \(P_{C_{3}H_{7}OH} = X_{C_{3}H_{7}OH} . P^0_{C_{3}H_{7}OH} = 0.700 . 37.6 mmHg = 26.32 mmHg\)

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