The vapour pressure of a solution of \(5 \mathrm{~g}\) of nonelectrolyte in \(100 \mathrm{~g}\) of water at a particular temperature is \(2985 \mathrm{Nm}^{-2}\). The vapour pressure of pure water at that temperature is \(3000 \mathrm{Nm}^{-2}\). The molecular weight of the solute is: (a) 180 (b) 90 (c) 270 (d) 360

where \(P\) is the vapor pressure of the solution, \(X_{solvent}\) is the mole fraction of the solvent, and \(P_0\) is the vapor pressure of the pure solvent. This relationship indicates that introducing a non-volatile solute, such as sugar or salt, into a solvent, like water, will lead to a reduction in the solution's vapor pressure. This reduction results because the solute particles take up space at the surface of the solution, thereby reducing the number of solvent particles that can escape into the vapor phase.

By understanding Raoult's Law, students can predict how changes in solution composition affect its physical properties. The law is crucial for problems involving colligative properties, like boiling point elevation, freezing point depression, and, as highlighted in our exercise, vapor pressure lowering.

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of moles of that component to the total number of moles of all components in the mixture. The mole fraction can be calculated using the formula:

\[ X_i = \frac{n_i}{\sum{n_i}} \]
, where \(X_i\) represents the mole fraction of component \(i\), \(n_i\) is the number of moles of component \(i\), and \(\sum{n_i}\) is the sum of the moles of all components.

In the context of the given exercise, the change in vapor pressure (\(\Delta P\)) between the pure solvent and the solution can be used to determine the mole fraction of the solute. This is because the change in vapor pressure is directly proportional to the mole fraction of the solute in accordance with Raoult's Law. By isolating the mole fraction in the equation representing Raoult's Law, students can use the given vapor pressure data to find the solute's mole fraction, which is a critical step in determining the molecular weight of the solute.

The determination of molecular weight, or molar mass, is a key process in many areas of chemistry and is required to convert between the mass of a substance and the number of moles. The molecular weight is the mass, in grams, of one mole of a substance. It is expressed in units of grams per mole (g/mol).

To determine the molecular weight from a given set of data as illustrated in our exercise, we must find the number of moles of the solute (\(n_s\)) using the mole fraction and the mass of solute. Once we know the number of moles and mass of the solute, we can rearrange the definition of molecular weight (\( M = \frac{mass}{moles}\)) to solve for the solute's molecular weight.

By applying the mole fraction found through vapor pressure lowering, along with the known mass of the solute and the solvent, we can construct an equation that relates all these factors. Solving this equation provides us with the molecular weight of the unknown solute, a vital piece of information for understanding the solution's composition and for various practical applications, like dosing of medications, formulation of mixtures, and in chemical synthesis.