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Chapter 11: Problem 114

A solution is prepared by mixing \(1.000\) mole of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) and \(3.18\) moles of propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right) .\) What is the composition of the vapor (in mole fractions) at \(40^{\circ} \mathrm{C}\) ? At \(40^{\circ} \mathrm{C}\), the vapor pressure of pure methanol is 303 torr, and the vapor pressure of pure propanol is \(44.6\) torr.

Short Answer

Expert verified
The composition of the vapor (in mole fractions) at 40°C is: Methanol: 0.6810 and Propanol: 0.3190.

Step by step solution

01

Calculate mole fractions of methanol and propanol in the solution

We will start by calculating the total moles of the solution, then find the mole fraction of each component. The total moles in the solution are: Total moles = moles of methanol + moles of propanol Total moles = 1.000 + 3.18 = 4.18 moles Now we can find the mole fractions: Mole fraction of methanol (X_methanol) = moles of methanol / total moles Mole fraction of propanol (X_propanol) = moles of propanol / total moles X_methanol = 1.000 / 4.18 = 0.2392 X_propanol = 3.18 / 4.18 = 0.7608
02

Apply Raoult's Law to find partial pressures of methanol and propanol in the vapor phase

Now, we'll use Raoult's Law to find the partial pressures of methanol and propanol in the vapor phase. P_methanol = X_methanol * vapor pressure of pure methanol P_propanol = X_propanol * vapor pressure of pure propanol At 40°C, the vapor pressure of pure methanol is 303 torr, and the vapor pressure of pure propanol is 44.6 torr. P_methanol = 0.2392 * 303 = 72.47 torr P_propanol = 0.7608 * 44.6 = 33.95 torr
03

Calculate mole fractions of methanol and propanol in the vapor phase

Finally, we will calculate the mole fractions in the vapor phase. To do this, we will first find the total vapor pressure: Total vapor pressure (P_total) = P_methanol + P_propanol P_total = 72.47 + 33.95 = 106.42 torr Now, we can find the mole fractions in the vapor phase: Y_methanol = P_methanol / P_total Y_propanol = P_propanol / P_total Y_methanol = 72.47 / 106.42 = 0.6810 Y_propanol = 33.95 / 106.42 = 0.3190 So, the composition of the vapor (in mole fractions) at 40°C is: Methanol: 0.6810 Propanol: 0.3190

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Most popular questions from this chapter

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is a monoprotic acid that ionizes only partially in aqueous solutions. A \(0.10-M\) formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of \(0.10 M\) formic acid.

Calculate the freezing point and boiling point of an antifreeze solution that is \(50.0 \%\) by mass of ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) in water. Ethylene glycol is a nonelectrolyte.

What is the composition of a methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) -propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) solution that has a vapor pressure of 174 torr at \(40^{\circ} \mathrm{C} ?\) At \(40^{\circ} \mathrm{C}\), the vapor pressures of pure methanol and pure propanol are 303 and \(44.6\) torr, respectively. Assume the solution is ideal.

A solid mixture contains \(\mathrm{MgCl}_{2}\) and NaCl. When \(0.5000 \mathrm{~g}\) of this solid is dissolved in enough water to form \(1.000 \mathrm{~L}\) of solution, the osmotic pressure at \(25.0^{\circ} \mathrm{C}\) is observed to be \(0.3950\) atm. What is the mass percent of \(\mathrm{MgCl}_{2}\) in the solid? (Assume ideal behavior for the solution.)

Benzene and toluene form an ideal solution. Consider a solution of benzene and toluene prepared at \(25^{\circ} \mathrm{C}\). Assuming the mole fractions of benzene and toluene in the vapor phase are equal, calculate the composition of the solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of benzene and toluene are 95 and 28 torr, respectively.
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