The vapor pressure of pure water at \(70^{\circ} \mathrm{C}\) is $31.2 \mathrm{kPa}\(. The vapor pressure of water over a solution at \)70^{\circ} \mathrm{C}$ containing equal numbers of moles of water and glycerol \(\left(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\right.\), a nonvolatile solute) is \(13.3 \mathrm{kPa}\). Is the solution ideal according to Raoult's law?

Short Answer

Expert verified
The given solution does not follow Raoult's law and is not an ideal solution because the expected vapor pressure of an ideal solution (\(15.6 \mathrm{kPa}\)) is different from the given vapor pressure of the solution (\(13.3 \mathrm{kPa}\)).

Step by step solution

01

Calculate the mole fraction of water

Given that the solution contains equal numbers of moles of water and glycerol, this means the mole fraction of water denoted as \(x_{water}\) can be calculated as follows: \[x_{water} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of glycerol}}\] Since they both have equal moles, we can denote the moles of water as \(n\) and moles of glycerol as \(n\): \[x_{water} = \frac{n}{n + n} = \frac{n}{2n} = \frac{1}{2}\]
02

Apply Raoult's law for an ideal solution

Now that we have the mole fraction of water, we'll apply Raoult's law for an ideal solution. Raoult's law states that: \[P_{solution}^{ideal} = x_{water} \times P_{water}\] Where \(P_{solution}^{ideal}\) is the expected vapor pressure of an ideal solution, \(x_{water}\) is the mole fraction of water, and \(P_{water}\) is the vapor pressure of pure water. Using the information given, we can calculate the expected vapor pressure of an ideal solution at \(70^{\circ} \mathrm{C}\): \[P_{solution}^{ideal} = \frac{1}{2} \times 31.2 \mathrm{kPa} = 15.6 \mathrm{kPa}\]
03

Compare the expected vapor pressure to the given vapor pressure

To determine if the given solution follows Raoult's law and behaves ideally, we need to compare the expected vapor pressure of an ideal solution, \(P_{solution}^{ideal}\), to the given vapor pressure of the solution, \(P_{solution}\): Given vapor pressure, \(P_{solution} = 13.3 \mathrm{kPa}\) Expected vapor pressure of an ideal solution, \(P_{solution}^{ideal} = 15.6 \mathrm{kPa}\) Since the expected vapor pressure of an ideal solution is different from the given vapor pressure of the solution, we can infer that the solution does not follow Raoult's law and is not an ideal solution.

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