Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution is obtained by dissolving \(6 \mathrm{~g}\) of urea (mol. wt \(=60\) ) in a litre solution, another solution is prepared by dissolving \(34.2 \mathrm{~g}\) of cane sugar (mol. \(\mathrm{wt}=342\) ) in a litre of solution at the same temperature The lowering of vapour pressure in the first solution is (a) same as that of second solution (b) double that of second solution (c) half that of second solution (d) nearly one fifth of the second solution

Step-by-Step Solution

Answer

(a) same as that of second solution

Step 1: Identify the formula

To find the lowering of vapor pressure, we use the formula for lowering of vapor pressure in a solution, given by \[\Delta P = P_0 \times x_{ ext{solute}}\]where \(P_0\) is the vapor pressure of the pure solvent and \(x_{ ext{solute}}\) is the mole fraction of the solute.

Step 2: Calculate mole fraction for urea solution

Calculate the number of moles of urea:\[\text{Moles of urea} = \frac{6}{60} = 0.1 \text{ moles}\]Assuming water is the solvent and is present as a litre, its mass is taken as 1000 g. The number of moles of water is:\[\text{Moles of water} = \frac{1000}{18} \approx 55.56 \text{ moles}\]The total moles in the solution are\[0.1 + 55.56 = 55.66 \text{ moles}\]Therefore, the mole fraction of urea is:\[x_{ ext{urea}} = \frac{0.1}{55.66} \approx 0.0018\]

Step 3: Calculate mole fraction for cane sugar solution

Calculate the number of moles of cane sugar:\[\text{Moles of cane sugar} = \frac{34.2}{342} = 0.1 \text{ moles}\]Using water as the solvent again, moles of water is\[55.56 \text{ moles}\]The total moles in the solution are:\[0.1 + 55.56 = 55.66 \text{ moles}\]So, the mole fraction of cane sugar is:\[x_{ ext{cane sugar}} = \frac{0.1}{55.66} \approx 0.0018\]

Step 4: Compare the results

Both solutions have the same number of moles of solute in a litre of solution, resulting in the same mole fraction of the solute. Thus, the lowering of vapor pressure is the same for both solutions.

Key Concepts

Mole Fraction

The concept of mole fraction is central to understanding how substances behave when dissolved in a solution. It is a way of expressing the concentration of a component in a mixture. More specifically, the mole fraction is the ratio of the number of moles of a particular component to the total number of moles in the solution. The formula for calculating mole fraction (x ext{solute}) is:\[x ext{solute} = \frac{ ext{moles of solute}}{ ext{total moles in solution}}\]This dimensionless quantity helps predict how a solute will affect the other properties of the solution, such as vapor pressure, boiling point, and freezing point. In the given exercise, both urea and cane sugar solutions have the same mole fraction of 0.0018. This means that the number of moles of solute per mole of solution is proportionally the same, which is crucial in comparing the effects each solute has on colligative properties like vapor pressure lowering.

Vapor Pressure Lowering

Vapor pressure lowering is an important colligative property, which means that it depends on the ratio of solute to solvent particles, regardless of the chemical identity of the solute. When a non-volatile solute is added to a solvent, the vapor pressure of the solution is lower than that of the pure solvent. This happens because the solute particles occupy space at the surface of the liquid, effectively reducing the number of solvent molecules that can escape into the vapor phase.
The formula used to determine vapor pressure lowering is:\[\Delta P = P_0 \times x\text{solute}\]Where \(\Delta P\) is the change in vapor pressure, \(P_0\) is the vapor pressure of the pure solvent, and \(x\text{solute}\) is the mole fraction of the solute. In the original problem, since the mole fraction of solute in both the urea and cane sugar solutions is identical, the lowering of vapor pressure is the same for both. This illustrates that vapor pressure lowering relies entirely on how many solute particles are in the solution, not what those particles are.

Solution Chemistry

Understanding solution chemistry is key to predicting and explaining the behavior of solutions, which are homogeneous mixtures of two or more substances. Solutions typically consist of a solute that is dissolved in a solvent. The properties of a solution depend not just on the nature of the solvent and solute, but also on the concentrations of all species involved.
A solution's colligative properties, such as vapor pressure lowering, boiling point elevation, and freezing point depression, are particularly interesting because they depend on the quantity, not the identity, of the solute present. These properties arise due to disruptions in the interactions between solvent molecules when solute molecules are added. In the given problem, urea and cane sugar, despite being different chemicals, produce the same effect on vapor pressure when dissolved in equal mole fractions. This illustrates a key point of solution chemistry: many properties of solutions depend on particle numbers rather than particle identities, making concepts like mole fraction crucial for understanding outcomes in solution chemistry.