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A solution is prepared by dissolving 45.2 g of a nonvolatile solute in 119 g of water. The vapor pressure above the solution is 22.51 Torr and the vapor pressure of pure water is 23.76 Torr at this temperature. What is the molecular weight of the solute?

Short Answer

Expert verified
To find the molecular weight of the solute, we first calculate the mole fraction of the solvent (water) using the given mass of solute and solvent. Next, apply Raoult's Law to find the molality of the solute in the solution. Then, calculate the moles of solute using the mass of the solvent (water). Finally, using the mass of the solute and the moles of the solute, we calculate the molecular weight of the solute using the formula: \(molecular\ weight = \frac{mass\ of\ solute}{moles\ of\ solute}\).

Step by step solution

01

Calculate the mole fraction of the solvent (water)

First, we need to find the mole fraction of the solvent (water). To do this, we will use the mass of the solute and the mass of the solvent (water) to determine the molality of the solution, and then we can find the mole fraction from the molality. Molality (m) of the solution is calculated as: \(m = \frac{moles\ of\ solute}{kg\ of\ solvent}\) The mole fraction (χ) of the solvent (water) can then be calculated as: \(\chi_{H_2O} = \frac{molality}{molality + 1}\)
02

Apply Raoult's Law to find the molality of the solute

Now that we have the mole fraction of the solvent (water), we can apply Raoult's Law to find the molality of the solute in the given solution. Raoult's Law states: \(P_{solution} = \chi_{H_2O}\times P_{H_2O}^{pure}\) Solve for molality: \(molality = \frac{P_{solution}}{P_{H_2O}^{pure} - P_{solution}} - 1\)
03

Calculate the moles of solute

Now that we've found the molality of the solute, we can calculate the moles of solute using the mass of the solvent (water). We know: \(m = \frac{moles\ of\ solute}{kg\ of\ solvent}\) So, we can solve for moles of solute: \(moles\ of\ solute = m \times kg\ of\ solvent\)
04

Calculate the molecular weight of the solute

Finally, we will use the mass of the solute and the moles of the solute to calculate the molecular weight of the solute. The formula to calculate the molecular weight is: \(molecular\ weight = \frac{mass\ of\ solute}{moles\ of\ solute}\) By plugging in the mass of the solute and the moles of solute calculated in the previous step, we can find the molecular weight of the solute.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Raoult's Law
Raoult's Law is an essential principle in chemistry that helps us understand how the addition of a solute affects the vapor pressure of a solvent. In simple terms, when a nonvolatile solute (a solute that does not evaporate) is added to a solvent, it lowers the solvent's vapor pressure. This happens because the solute molecules interrupt the evaporation process of the solvent at the surface, reducing the number of molecules that can escape into the vapor phase. The law is expressed as:
\[ P_{solution} = \chi_{solvent} \times P_{solvent}^{pure} \]
Here, \( P_{solution} \) is the vapor pressure of the solution, \( \chi_{solvent} \) is the mole fraction of the solvent, and \( P_{solvent}^{pure} \) is the vapor pressure of the pure solvent. Using this equation, we can calculate how much the vapor pressure decreases when a solute is added, which is crucial in determining properties such as boiling and freezing points.
  • Raoult's Law only applies to ideal solutions where interactions between molecules are similar to those in the pure components.
  • It's important when studying colligative properties of solutions.
Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. It's the ratio of the number of moles of one component to the total number of moles in the solution. The sum of the mole fractions in a mixture always equals one. In equations, the mole fraction (\( \chi \)) of a component A is given by:
\[ \chi_A = \frac{n_A}{n_{total}} \]
Where \( n_A \) is the number of moles of component A and \( n_{total} \) is the total number of moles in the solution.
  • It helps in calculating other properties such as vapor pressure and boiling point elevation.
  • Mole fraction is a dimensionless quantity, meaning it has no units.
  • This method is particularly useful in chemical bonding and reaction chemistry.
Molality
Molality is another way to express the concentration of a solution, particularly useful when dealing with temperature variations. It measures the number of moles of solute per kilogram of solvent, not the overall solution. The formula is given by:
\[ m = \frac{ moles\ of\ solute }{ kg\ of\ solvent } \]
Unlike molarity, molality does not change with temperature since it's based on mass, which remains constant. This makes it a preferred unit in scenarios where temperature changes are involved.
  • It is vital for calculating colligative properties, which depend on the amount of solute, not its identity.
  • Molality is most useful in laboratory settings where precise concentration measurements are required.
  • Because it is based on mass, it is unaffected by changes in physical states (solid, liquid, gas).
Nonvolatile Solute
A nonvolatile solute is a substance that does not evaporate into a gas under existing conditions. When added to a solvent, it causes a change in the physical properties of the solution, like boiling point elevation or freezing point depression, by affecting the liquid-vapor equilibrium.
  • These solutes do not contribute to the vapor above the solution, so their presence lowers the solution's vapor pressure.
  • Common examples are salt and sugar when dissolved in water.
  • They are key in studying and applying Raoult's Law, as they directly impact the solution's behavior.
They are critical to many industrial and scientific applications, especially in processes such as freeze-drying, where controlling moisture content is essential.

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Most popular questions from this chapter

At 303. K, the vapor pressure of benzene is 120. Torr and that of hexane is 189 Torr. Calculate the vapor pressure of a solution for which \(x_{\text {benzene }}=0.28\) assuming ideal behavior.

The data from Problem P9.20 can be expressed in terms of the molality rather than the mole fraction of \(\mathrm{Br}_{2}\). Use the data from the following table and a graphical method to determine the Henry's law constant for \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\) at \(25^{\circ} \mathrm{C}\) in terms of molality. $$\begin{array}{lllc} m_{B r_{2}} & P(\text { Torr }) & m_{B r_{2}} & P(\text { Torr }) \\ \hline 0.026 & 1.52 & 0.086 & 5.43 \\ 0.028 & 1.60 & 0.157 & 9.57 \\ 0.039 & 2.39 & 0.158 & 9.83 \\ 0.067 & 4.27 & 0.167 & 10.27 \end{array}$$

A and \(B\) form an ideal solution. At a total pressure of \(0.720 \mathrm{bar}, y_{A}=0.510\) and \(x_{A}=0.420 .\) Using this information, calculate the vapor pressure of pure \(A\) and of pure \(B\).

An ideal solution is made up of the volatile liquids A and \(\mathrm{B}\), for which \(P_{A}^{*}=165\) Torr and \(P_{B}^{*}=85.1\) Torr. As the pressure is reduced, the first vapor is observed at a total pressure of \(110 .\) Torr. Calculate \(x_{\mathrm{A}}\).

The dissolution of 7.75 g of a substance in 825 g of benzene at \(298 \mathrm{K}\) raises the boiling point by \(0.575^{\circ} \mathrm{C}\). Note that \(K_{f}=5.12 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}, K_{b}=2.53 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1},\) and the density of benzene is \(876.6 \mathrm{kg} \mathrm{m}^{-3} .\) Calculate the freezing point depression, the ratio of the vapor pressure above the solution to that of the pure solvent, the osmotic pressure, and the molecular weight of the solute. \(P_{\text {benzene }}^{*}=103\) Torr at \(298 \mathrm{K}\).

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