Question

Vapour pressure of a solution of \(5 \mathrm{~g}\) of non-electrolyte in \(100 \mathrm{~g}\) of water at \(\mathrm{a}\) particular temperature is \(2985 \mathrm{~N} / \mathrm{m}^{2}\) If the vapour pressure of pure water at this temperature is \(3000 \mathrm{~N} / \mathrm{m}^{2}\), the molecular mass of the solute is (a) 60 (b) 120 (c) 180 (d) 360

Short answer

The molecular mass of the solute is 180 g/mol.

Step-by-step solution

Understand and Apply Raoult's Law

Raoult's law states that the vapour pressure of a solvent above a solution is equal to the vapour pressure of the pure solvent at the same temperature, multiplied by the mole fraction of the solvent in the solution. If we denote the vapour pressure of the solution as P_solution, the vapour pressure of the pure solvent as P_pure, and the mole fraction of the solvent as X_solvent, Raoult's law can be written as: \( P_{solution} = X_{solvent} \times P_{pure} \). Since we're given P_solution and P_pure, we can use this relation to find the mole fraction of the solvent.

Calculate Mole Fraction of Solvent

To find the mole fraction of the solvent (water), we first need to calculate the vapour pressure decrease, which is \( P_{pure} - P_{solution} = 3000 \, \mathrm{N/m^2} - 2985 \, \mathrm{N/m^2} = 15 \, \mathrm{N/m^2} \). To get the mole fraction of the solvent, we then divide the vapour pressure decrease by the vapour pressure of the pure solvent: \( X_{solvent} = \frac{15}{3000} = 0.005 \).

Calculate Mole Fraction of Solute

The mole fraction of the solute (X_solute) can be found using the relation \( X_{solute} = 1 - X_{solvent} \). Substituting the value obtained in Step 2 gives us: \( X_{solute} = 1 - 0.005 = 0.995 \).

Calculate Moles of Water and Solute

The amount of water in moles (n_water) can be calculated using its mass (100 g) and its molar mass (18.015 g/mol): \( n_{water} = \frac{100 g}{18.015 g/mol} \). The amount of solute in moles (n_solute) can then be calculated from the mole fraction (using the relation \( X_{solute} = \frac{n_{solute}}{n_{solute} + n_{water}} \)): \( n_{solute} = \frac{X_{solute} }{1- X_{solute} } \times n_{water} \).

Determine Molecular Mass of the Solute

We can find the molecular mass (M_solute) of solute using its given mass (5 g) and the moles (n_solute) calculated in Step 4: \( M_{solute} = \frac{mass_{solute}}{n_{solute}} \).

Calculate and Identify the Correct Answer

Now plug the values into the molecular mass formula from Step 5 and calculate the molecular mass of the solute. Identify the correct answer among the given options.

Key concepts

Vapour Pressure

Vapour pressure is a critical concept in understanding the properties of liquids and solutions. It refers to the pressure exerted by the vapour that is in equilibrium with its liquid (or solid) at a given temperature. In simpler terms, it's the force at which molecules escape from a liquid to become gas.

Now, why does this matter in solutions? When a non-electrolyte solute is dissolved in a solvent like water, it decreases the tendency of the water molecules to escape into the vapour phase, which in turn lowers the vapour pressure of the solution compared to the pure solvent at the same temperature. This phenomenon underpins Raoult's Law and is critical to understanding how solutions behave.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of that component to the total number of moles of all components in the mixture.

For example, if we're looking at a saline solution made of salt and water, we can calculate the mole fraction of each component. The mole fraction is an important part of Raoult's Law, as it helps determine the vapour pressure of the solution based on the proportion of solvent and solute present. Remember, in a binary solution (two-part mixture), the sum of the mole fractions of both components is always equal to one.

Molecular Mass

Molecular mass (also known as molecular weight) is the mass of a single molecule of a substance. It is typically expressed in atomic mass units (amu) or g/mol. In practical terms, molecular mass can be calculated by adding together the atomic masses of the atoms in the molecule. Knowing the molecular mass is crucial when you're converting between mass in grams and the amount of substance in moles – a key step in many chemical calculations, including those involving Raoult's Law.

To solve for the unknown molecular mass of a solute in a solution, you can rearrange the mole fraction formula after determining the moles of solute and solvent present. This process requires a comprehensive understanding of stoichiometry and the use of molecular mass as a conversion factor.

Non-Electrolyte Solutions

Finally, let's talk about non-electrolyte solutions. These solutions contain a solute that, when dissolved in a solvent, does not dissociate into ions. While electrolyte solutions conduct electricity due to the presence of ions, non-electrolyte solutions do not. This distinction is important when applying Raoult's Law because it only holds true for ideal solutions – typically non-electrolyte solutions with similar molecular interactions between solute and solvent molecules.

In the context of Raoult's Law problems, non-electrolyte solutions are preferred because they do not introduce the complexity of ionic interactions, which can affect vapour pressure and lead to deviations from the law. When dealing with non-electrolyte solutions, it is assumed that the solute simply dilutes the solvent, causing the vapour pressure to decrease proportionally to the amount of solute added.