Question

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C} )\) b. a solution of glucose in water with \(\chi_{\mathrm{CH}_{2} \mathrm{O}_{6}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH}, 0 \mathrm{H}}=0.2\) (Consider the vapor pressure of both methanol \([143 \text { torr at }\) \(25^{\circ} \mathrm{C} ]\) and water.

Short answer

The solution with the lowest total vapor pressure at \(25^{\circ} \mathrm{C}\) is the glucose solution (b) with a vapor pressure of \(23.562 \mathrm{torr}\).

Step-by-step solution

Pure water vapor pressure

For pure water, the vapor pressure is already given as \(23.8\) torr. So, no further calculation is needed for this case.

Glucose solution vapor pressure

Glucose is a non-volatile solute, which means it doesn't contribute to the vapor pressure. Using Raoult's law, we can find the vapor pressure of water in this solution: \[P_{\text{H}_2\text{O}} = \chi_{\text{H}_2\text{O}} * P_{\text{pure H}_2\text{O}}\] Since mole fraction of glucose (\(\chi_{\mathrm{CH}_{2}\mathrm{O}_{6}}\)) is \(0.01\), mole fraction of water is \(1 - 0.01 = 0.99\). Therefore, \[P_{\text{H}_2\text{O}} = 0.99 * 23.8 \mathrm{torr} = 23.562 \mathrm{torr}\]

Sodium chloride solution vapor pressure

Since NaCl is an ionic compound, its dissolution into water will increase the number of solute particles, forming 2 ions (Na\(^+\) and Cl\(^-\)) for each NaCl formula unit. This will affect the vapor pressure, which does not follow Raoult's law ideally: \[P_{\text{NaCl}} = i * \chi_{\text{H}_2\text{O}} * P_{\text{pure H}_2\text{O}}\] where \(i\) is the van't Hoff factor, in this case \(i=2\). For this case, \(\chi_{\mathrm{NaCl}}=0.01\), thus mole fraction of water is \(1-0.01=0.99\). Then, \[P_{\text{H}_2\text{O}} = 2 * 0.99 * 23.8 \mathrm{torr} = 47.124 \mathrm{torr}\]

Methanol solution vapor pressure

For the methanol-water solution, we have two volatile components, which means we need to take into account both components' vapor pressure. Applying Raoult's law for each component: \[P_{\text{H}_2\text{O}} = \chi_{\text{H}_2\text{O}} * P_{\text{pure H}_2\text{O}}\] \[P_{\text{CH}_3\text{OH}} = \chi_{\text{CH}_3\text{OH}} * P_{\text{pure CH}_3\text{OH}}\] Here, \(\chi_{\mathrm{CH}, 0 \mathrm{H}}=0.2\), thus mole fraction of water is \(1-0.2=0.8\). Then, \[P_{\text{H}_2\text{O}} = 0.8 * 23.8 \mathrm{torr} = 19.04 \mathrm{torr}\] \[P_{\text{CH}_3\text{OH}} = 0.2 * 143 \mathrm{torr}= 28.6 \mathrm{torr}\] The total vapor pressure of the methanol solution is the sum of the partial pressures of water and methanol. \[P_{\text{total}} = P_{\text{H}_2\text{O}} + P_{\text{CH}_3\text{OH}} = 19.04 \mathrm{torr} + 28.6 \mathrm{torr} = 47.64 \mathrm{torr}\]

Comparing vapor pressures

Now, let's compare the vapor pressures obtained for each case: Pure water: \(23.8 \mathrm{torr}\) Glucose solution: \(23.562 \mathrm{torr}\) NaCl solution: \(47.124 \mathrm{torr}\) Methanol solution: \(47.64 \mathrm{torr}\) The solution with the lowest total vapor pressure is the glucose solution (b).

Key concepts

Raoult's Law

Raoult's Law gives us a way to predict the vapor pressure of a solution. It's especially useful when dealing with solutions of non-volatile solutes, like glucose. According to Raoult's Law, the vapor pressure of a solvent above a solution (\(P_{\text{solution}}\)) is equal to the mole fraction of the solvent in the solution (\(\chi_{\text{solvent}}\)) multiplied by the vapor pressure of the pure solvent (\(P_{\text{pure solvent}}\)). This is mathematically expressed as:

\[P_{\text{solution}} = \chi_{\text{solvent}} \times P_{\text{pure solvent}}\]

This means that when you dissolve a non-volatile solute in a solvent, the solution's vapor pressure will be lower than the vapor pressure of the pure solvent. Why does this happen? The presence of solute molecules reduces the number of solvent molecules at the surface, thereby decreasing the vapor pressure.

So, when we say a solution "obeys Raoult's Law," it means this relationship accurately describes its behavior.

Mole Fraction

The mole fraction helps us understand how abundant a particular component is in a mixture. It's not the amount by mass or volume but rather the amount by moles, which is a common way to quantify chemical amounts.

To calculate the mole fraction (\(\chi\)) of a substance, use the formula:

\[\chi = \frac{\text{moles of component}}{\text{total moles of all components}}\]

The sum of the mole fractions of all components in a mixture is always equal to 1. It's a crucial piece of information because, in Raoult's Law, the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. That's why, for example, knowing the mole fraction of glucose is helpful in determining the vapor pressure of a glucose-water solution.

Non-volatile Solute

A non-volatile solute, such as glucose, is a solute that does not easily vaporize at the temperature of interest. Because of this, a solution containing a non-volatile solute will have a lower vapor pressure compared to the pure solvent.

Non-volatile solutes don’t contribute to the vapor pressure above the solution. In such a case, Raoult’s Law applies nicely and allows us to directly relate the lowering of the vapor pressure to the mole fraction of the solvent.

The presence of the non-volatile solute in the solution means that there are fewer solvent molecules at the surface available to escape into the vapor phase, which reduces the vapor pressure. Hence, in scenarios where only the solvent contributes to the vapor, the vapor pressure decreases, illustrating what's known as 'vapor pressure lowering,' a colligative property.

Van't Hoff Factor

The van't Hoff factor \((i)\) is an important concept when dealing with ionic solutes like sodium chloride \((\text{NaCl})\). It accounts for the number of particles an ionic compound dissociates into when it dissolves in a solvent.

For non-ionic substances like glucose, \(i\) is typically 1, because the compound does not dissociate in solution. However, for \(\text{NaCl}\), the van't Hoff factor is 2 because it splits into two ions: \(\text{Na}^+\) and \(\text{Cl}^-\).

This factor is essential because it modifies how we apply Raoult's Law and calculate colligative properties, such as vapor pressure lowering, freezing point depression, and boiling point elevation.

• With the van't Hoff factor, we consider how many particles are in the solution, which then affects the properties of the solution.

Understanding \(i\) helps predict how different solutes will alter the solutions' properties, thus providing insight into the behavior of ionic versus molecular solutes.