Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

The mass of glucose that should be dissolved in \(50 \mathrm{~g}\) of water in order to produce the same lowering of vapour pressure as is produced by dissolving \(1 \mathrm{~g}\) of urea in the same quantity of water is (a) \(1 \mathrm{~g}\) (b) \(3 \mathrm{~g}\) (c) \(6 \mathrm{~g}\) (d) \(18 \mathrm{~g}\)

Step-by-Step Solution

Answer

The mass of glucose needed is approximately \(3 \text{ g}\), option (b).

Step 1: Understand Raoult's Law

Raoult's law states that the relative lowering of vapor pressure of a solvent by a non-volatile solute is equal to the mole fraction of the solute in the solution. Mathematically, this is expressed as: \[ \text{Relative lowering of vapor pressure} = \frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}} \] For dilute solutions, this simplifies to: \[ \frac{\Delta P}{P_0} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} \] where \( n_{\text{solute}} \) is the number of moles of solute and \( n_{\text{solvent}} \) is the number of moles of solvent.

Step 2: Calculate Moles of Urea

Urea has a molar mass of \(60 \text{ g/mol}\). For \(1 \text{ g}\) of urea dissolved in water, the moles of urea can be calculated as:\[ n_{\text{urea}} = \frac{1 \text{ g}}{60 \text{ g/mol}} = 0.0167 \text{ mol} \]

Step 3: Calculate Moles of Water

The molar mass of water \( \text{H}_2\text{O} \) is \(18 \text{ g/mol}\). For \(50 \text{ g}\) of water, the moles are calculated as:\[ n_{\text{water}} = \frac{50 \text{ g}}{18 \text{ g/mol}} = 2.78 \text{ mol} \]

Step 4: Setup the Equation for Glucose

Let the mass of glucose dissolved be \(x \text{ g}\), and its molar mass is \(180 \text{ g/mol}\). The moles of glucose then is:\[ n_{\text{glucose}} = \frac{x}{180} \]According to the question, the relative lowering of vapor pressure by glucose should match that of urea:\[ \frac{0.0167}{2.78} = \frac{\frac{x}{180}}{2.78} \]

Step 5: Solve for the Mass of Glucose

Equate and solve for \(x\):\[ 0.0167 = \frac{x}{180} \]Multiplying both sides by 180 gives:\[ x = 180 \times 0.0167 = 3.006 \text{ g} \]Hence, approximately \(3 \text{ g}\) of glucose is needed.

Key Concepts

Vapor Pressure Lowering

The concept of vapor pressure lowering is a fundamental principle in chemistry, especially when dealing with solutions. When a non-volatile solute like glucose or urea is added to a solvent like water, it causes the vapor pressure of the solvent to decrease. This is because solute particles occupy space at the liquid's surface, reducing the number of solvent molecules that can escape into the vapor phase. This phenomenon is crucial in understanding how solutions behave and is quantitatively described by Raoult's Law.

Raoult's Law states that the vapor pressure of a solvent in a solution is proportional to the mole fraction of the solvent. When a solute is added, it disrupts this balance, resulting in a decrease in vapor pressure. The relative lowering of vapor pressure compared to the pure solvent can be calculated using the formula: \[ \frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}} \]For dilute solutions, this simplifies to:\[ \frac{\Delta P}{P_0} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]

Mole Fraction

Mole fraction is a way of expressing concentration, represented by the ratio of the number of moles of a component to the total number of moles in the mixture. It has no units and provides insight into the composition of mixtures. In the context of Raoult's Law, the mole fraction of the solute is crucial for determining how much the vapor pressure is lowered.

For example, in the exercise, we calculate the mole fraction of urea, given its mass and molar mass. By knowing the solvent's mass and molar mass, one can compute the mole fraction of each component, which helps set up equations that fulfill Raoult's Law's requirements. This allows for precise calculations of quantities required to achieve specific vapor pressure changes.

Molar Mass Calculation

Understanding molar mass calculation is vital in applications of Raoult's Law. Molar mass is defined as the mass of one mole of a substance, typically measured in grams per mole (g/mol). This metric is essential in converting between the mass of a substance and the amount of substance in moles.

In the exercise, molar mass calculations are used to determine the number of moles of glucose and urea. For instance, knowing that glucose has a molar mass of 180 g/mol and urea has a molar mass of 60 g/mol helps us compare their effects on vapor pressure lowering when dissolved in a common solvent. These calculations set the stage for determining how much of a substance is needed to produce a certain effect in a solution.

Glucose and Urea Comparison

Comparing substances like glucose and urea in terms of their effect on vapor pressure lowering is crucial to understanding their behavior in solutions. Both are non-volatile solutes and work similarly to lower the vapor pressure of a solvent through their presence in solution.

To produce the same effect on vapor pressure, the number of moles of solute is more important than the type of solute. Since glucose has a higher molar mass than urea, more mass of glucose is needed to achieve the same number of moles as urea. In the exercise, we find that 3 g of glucose—which roughly equates to 0.0167 moles—creates the same vapor pressure lowering effect as 1 g of urea. This type of practical comparison aids in predicting how different solutes behave in similar conditions.